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~LI\~rbite 71Pll7 NOTRE DAME MATHEMATICAL LECTURES Number 2 GALOIS THEORY Lectures delivered at the University of Notre Dame by DR. EMIL ARTIN Professor of Mathematics, Princeton University Edited and supplemented with a Section on Applications by DR. ARTHUR N. MILGRAM Associate Professor of Mathematics, University of Minnesota Second Edition With Additions and Revisions UNIVERSITY OF NOTRE DAME NOTRE DAME PRESS LONDON Copyright 1942, 1944 UNNERSITY OF NOTRE DAME Second Printing, February 1964 Third Printing, July 1965 Fourth Printing, August 1966 New composition with corrections Fifth Printing, March 1970 Sixth Printing, January 197 1 Printed in the United States of America by NAPCO Graphic Arts, Inc., Milwaukee, Wisconsin TABLE OF CONTENTS (The sections marked with an asterisk have been herein added to the content of the first edition) Page LINEAR A ALGEBRA ................................... . Fields ......................................... . Vector Spaces ................................... . C. Homogeneous Linear Equations . . . . . . . . . . . . . . . . . . . . . D. Dependence and Independence of Vectors .. , . . . . . . . . . E. Non-homogeneous Linear Equations . . . . . . . . . . . . . . . . . F.* Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B. II FIELD THEORY ............................ . A. B. C. D. E. Extension Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Algebraic Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Unique Decomposition of Polynomials into hreducible Factors . . . . . . . . . . . . . . . . . . . . . F. Group Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G.* Applications and Examples to Theorem 13 . . . . . . . . . . . . H, Nmmal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I. Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ]. Roots of Unity............................. .., .. K. Noether Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . L Kummer's Fields . . . . . . . . . . . . . . . . . . . . . . . .......... M. Simple Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N. Existence of a Normal Basis . . . . . . . . . . . , . . . . . . . . . . . Q, Theorem on Natural hrationalities . . . . . . . . . . . . . . . . . . . Ill APPLICATIONS By A. N. Milgram •...................... A. Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B. Permutation Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C. Solution of Equations by Radicals . . . . . . . . . . . . . . . . . . . D. The General Equation of Degree n. . . . . . . . . . . . . . . . . . . E. Solvable Equations of Prime Degree . . . . . . . . . . . . . . . . . F. Ruler and Compass Construction . . . . . . . . . . . . . . . . . . . . 2 4 9 11 21 21 22 25 30 33 34 38 41 49 56 57 59 64 66 67 69 69 70 72 74 76 80 1 LINEAR ALGEBRA A. FVildB. (ft·; c.O\· f S) A field is a set of elements in which a pair of operations called multiplication and addition is defined analogous to the operations of multiplication and addition in the real number system (which is itself an example of a field). In each field F there exist unique elements called o and 1 which, under the operations of addition and multiplica- tion, behave with respect to all the other elements ofF exactly as their correspondents in the real number system. In two respects, the analogy is not complete: 1) multiplication is not assumed to be commu- tative in every field, and 2) a field may have only a finite number of elements. More exactly, a field is a set of elements which, under the above mentioned operation of addition, forms an additive abelian group and for which the elements, exclusive of zero, form a multiplicative group and, finally, in which the two group operations are connected by the distributive law. Furthermore, the product of o and any element is defined to be o. If multiplication in the field is commutative, then the field is called a commutative field. B. Vector Spaces. If Vis an additive abelian group with elements A, B, ... , F a field with elements a, b, ... , and if for each a f F and A f V 2 the product aA denotes an element of V, then V is called a (left) vector space over F if the following assumptions hold: l)a(A+B)=aA+aB 2) (a + b)A = aA + bA 3) a(bA) = ( ab )A 4) lA = A The reader may readily verify that if V is a vector space over F, then oA = 0 and aO = 0 where o is the zero element of F and 0 that of V. For example, the first relation follows from the equations: aA = (a+ o)A = aA + oA Sometimes products between elements ofF and V are written in the form A a in which case V is called a right vector space over F to distinguish it from the previous case where multiplication by field ele~ ments is from the left. If, in the discussion, left and right vector spaces do not occur simultaneously, we shall simply use the term "vector space." C. Homogeneous Linear Equations. If in afield F, aij, i = 1,2, ... , m,j = 1,2, ... , n arem. n ele- ments, it is frequently necessary to know conditions guaranteeing the existence of elements in F such that the following equations are satisfied: (1) am 1x 1 + am2x 2 + . . . + amnxn = 0. The reader will recall that such equations are called linear homogeneous equations, and a set of elements, X , 1 x2 , , .. , X of F, for which all the above equations are true, is called 0 a solution of the system. If not all of the elements x 1 , x , . xn 2 are o the solution is called non-trivial; otherwise, it is called trivial. THEOREM 1. A system of linear homogeneous equations always has a non-trivial solution if the number of unknowns exceeds the number of equations. The proof of this follows the method familiar to most high school students, namely, successive elimination of unknowns. If no equations in n > 0 variables are prescribed, then our unknowns are unrestricted and we may set them all = 1. We shall proceed by complete induction. Let us suppose that each system of k equations in more than k unknowns has a non-trivial solution when k < m. In the system of equations (1) we assume that n > m, and denote the expression ailx 1 + We seek elements x , . 1 = , xn not all o such that L = L = 1 2 If a 1J ~ o for each i and j, then any choice of x , 1 L m . , xn will serve as a solution. If not all aij are o, then we may assume that a 11 f o, for the order in which the equations are written or in which the unknowns are numbered has no influence on the existence or non-existence of a simultaneous solution. We can find a non-trivial solution to our given system of equations, if and only if we can find a non-trivial solution to the following system: L1 = o L2- a21ai1-1Lt = 0 0. For, if x 1 , L1 , . , xn is a solution of these latter equations then, since = o, the second term in each of the remaining equations is o and, Lm = o. Conversely, if (1) is satisfied, then the new system is clearly satisfied. The reader will notice that the new system was set up in such a way as to "eliminate" x from the 1 last m-1 equations. Furthermore, if a non-trivial solution of the last m-1 equations, when viewed as equations in x ,. 2 , xn, exists then + ~n Xn) would give us a solution to the whole system. However, the last m-1 equations have a solution by our inductive assumption, from which the theorem follows. Remark: If the linear homogeneous equations had been written in the form ~xja ij:::: o, j :::: 1, 2, , n, the above theorem would still hold and with the same proof although with the order in which terms are written changed in a few instances. D. Dependence and Independence of Vectors. In a vector space V over a field F, the vectors A, called dependent if there exist elements x , . 1 , xn, not , A n are all o, ofF such + xnAn = 0. If the vectors A, not dependent, they are called independent. The dimension of a vector space V over a field F is the maximum number of independent elements in V. Thus, the dimension of V is n if there are n independent elements in V, but no set of more than n independent elements. A system A, , A, of elements in V is called a generating system of V if each element A of V can be expressed 5 m Am, i.e., A = :£a1.A, for a suitable choice linearly in terms of A, i ~ 1 m,inF. of ai, i = 1,. THEOREM 2. In any generating system the maximum number of independent vectors is equal to the dimension of the vector space . Let A, . , A,, be a generating system of a vector space V of dimension n. Let r be the maximum number of independent elements in the generating system. By a suitable reordering of the generators we may as- sume A , . . . , Ar independent. By the definition of dimension it follows that 1 r < n. For each j, A, expressing this, a of A, r+i A,. f A . are dependent, and in the relation "' o, for the contrary would assert the dependence , Ar. Thus, It follows that A, , Ar is also a generating system since in the linear relation for any element of V the terms involving Ar+i, j I: o, can all be replaced by linear expressions in A, Now, let B 1 , ... , Bt be any system of vectors in V where t > r, then there exist a .. such that B."""" L a .. A., j = 1, 2, lJ J i"""" l lj , t, since the 1 A/ s form a generating system. If we can show that B, dependent, this will give us r _2 n, and the theorem will follow from this together with the previous inequality r < n~ Thus, we must ex- hibit the existence of a non-trivial solution out of F of the equation 0. 6 'lb this end, it will be sufficient to choose the x.'s so as to satisfy t ' the linear equations I x. a .. = o, i = 1, 2, ... , r, since these ex- i=l J t lJ pressions will be the coefficients of A 1 when in I x. B. the BJ 's are r j==tJJ replac;d by i~I a ij A 1 and terms are collected. A solution to the equations I x.a .. J . i=l = 0, i = 1,2, . . . , r, always exists by Theorem 1. lj Remark: Any n independent vectors A, ... , A, in an n dimensional vector space form a generating system. For any vector A, the vectors A, A, ... , A, are dependent and the coefficient of A, in the dependence relation, cannot be zero. Solving for A in terms of A 1' ... , A, exhibits A, ... ,An as a generating system. A subset of a vector space is called a subspace if it is a sub- group of the vector space and if, in addition, the multiplication of any element in the subset by any element of the field is also in the subset. If Ai, ... , As are elements of a vector space V, then the set of allele- ments of the form a, A, + ... + as As clearly forms a subspace of V. It is also evident, from the definition of dimension, that the dimension of any subspace never exceeds the dimension of the whole vector space. An s-tuple of elements ( a,, ... , a.) in a field F will be called a _row vecto_r. The totality of such s-tuples form a vector space if we define a) ( a 1 , a2 , ••• , a.)= ( b 1 , h 2 , ••• , b,) if and only if a, = b 1' i = 1,... ' s, {3) (a 1 ,a 2 , ... ,a,) 1 (b 1 ,h 2 , ... ,h,) y) b(a 1 ,a 2 , ••• ,a,)= (ba 1 ,ba 2 , ••• , ba,), for ban element of F. w"~ '"" ''"'~' "" ""'~ -"""''() they will be called column vectors. THEOREM 3. The row (column) vector space Fn of all n-tuples from a field F is a vector space of dimension n over F. The n elements (J ,, = ( o, 1, (n 1 Q) 0 ' .•• ' 0) . . . , 0, 1) = ( 11 0, 0 = ( O, 0, 1 • , are independent and generate Fn. Both remarks follow from the relation (al,a2, ... ,an)= 2ai(i, We call a rectangular anay amt 8 m2'' ' 8 mn of elements of a field F a matrix. By the right row rank of a matdx, we mean the maximum number of independent row vectors among the rows (ail',,., a,) of the matrix when multiplication by field elements is from the right. Similarly, we define left row rank, light column rank and left column rank. THEOREM 4. In any matrix the right column rank equals the left row rank and the left column rank equals the right row rank. If the field 8 is commutative, these four numbers are equal to each other and are called the rank of the matrix. Call the column vectors of the matrix Cl'. vectors R 1 , . . , Rm. The column vector 0 is dependence C 1 x 1 + C 2 x 2 + . , Cn and the row and any 0 is equivalent to a solution of the equations 0 (1) Any change in the order in which the rows of the matrix are written gives rise to the same system of equations and, hence, does not change the column rank of the matrix, but also does not change the row rank since the changed matrix would have the same set of row vectors. Call c the right column rank and r the left row rank of the matrix. By the above remarks we may assume that the first r rows are independent row vectors. The row vector space generated by all the rows of the matrix has, by Theorem 1, the dimension r and is even generated by the first r rows. Thus, each row after the the first r rows. Consequently, rth IS linearly expressible in terms of any solution of the first r equations in (1) will be a solution of the entire system since any of the last n-r equations is obtainable as a linear combination of the first r. Con- versely, any solution of (1) will also be a solution of the first r equations. This means that the matrix 9 allal2. ·a,n) ( 8 rt 8 r2 · arn consisting of the first r rows of the original matrix has the same right column rank as the original. It has also the same left row rank since the r rows were chosen independent. But the column rank of the amputated matrix car-mot exceed r by Theorem 3. Hence, c calling c' ~ r. Similarly, the left column rank and r' the right row rank, c' <... r'. If we form the transpose of the original matrix, that is, replace rows by columns and columns by rows, then the left row rank of the transposed matrix equals the left column rank of the original. If then to the transposed matrix we apply the above considerations we arrive at r < c and r' < c'. E. Non-homogeneous Linear Equations. The system of non-homogeneous linear equations 3 11 xl + 3 12x2 + a21 xl + + 8 1nxn = bl + a2nxn = b2 (2) has a solution if and only if the column vector in the space generated by the vectors cr Cf(U 10 This means that there is a solution if and only if the right column rank of the matrix ( a 11 ..• a 1" ) is the same as the amt· .. amw right column rank of the augmented matrix ( : ' · ~n b1 ) ami* '' 8 mnbm since the vector space generated by the original must be the same as the vector space generated by the augmented matrix and in either case the dimension is the same as the rank of the matrix by Theorem 2. By Theorem 4, this means that the row tanks are equal. Conversely, if the row rank of the augmented matrix is the same as the row rank of the original matdx, the column ranks will be the same and the equations will have a solution. If the equations (2) have a solution, then any relation among the rows of the original matrix subsists among the rows of the augmented matrix. For equations (2) this merely means that like combinations of equals are equal. Conversely, if each relation which subsists be- tween the rows of the original matrix also subsists between the rows of the augmented matrix, then the row rank of the augmented matrix is the same as the row rank of the original matrix. In terms of the equations this means that there will exist a solution if and only if the equations are consistent, i.e., if and only if any dependence between the left hand sides of the equations also holds between the light sides. 11 THEOREM 5. If in equations (2) m = n, there exists a unique solution if and only if the conesponding homogeneous equations allxt + ~2x2 + . . . + alnxn 8 n1 xl + an2x2 + . . . + 3 nnxn = 0 ° have only the trivial solution. If they have only the trivial solution, then the column vectors are independent. It follows that the original n equations in n unknowns will have a unique solution if they have any solution, since the difference, term by term, of two distinct solutions would be a non-trivial solution of the homogeneous equations. A solution would exist since the n independent column vectors form a generating system for the n-dimensional space of column vectors. Conversely, let us suppose our equations have one and only one solution. In this case, the homogeneous equations added term by te1m to a solution of the original equations would yield a new solu- tion to the original equations. Hence, the homogeneous equations have only the trivial solution. F. Detenninants. n The theory of determinants that we shall develop in this chapter is not needed in Galois theory. The reader may 1 therefore, omit this section if he so desires. We assume our field to be c o m m ut a t i v e and consider the square matrix ------~~--~~----~~~~~----- 1) Of the preceding theory only Theorem 1, for homogeneous equations and the notion of linear dependence are assumed known. 12 (1) ani an2 . . . . 3 nn of n rows and n columns. We shall define a certain function of this matrix whose value is an element of our field. The function will be called the determinant and will be denoted by 8..2ta22···· 3 2n (2) '"'''''"" or by D ( A , A ,. , , An) if we wish to consider it as a function of the 1 2 column vectors A, A, . A, of (1). If we keep all the columns but A, constant and consider the determinant as a function of A, then we write Dk( Ak) and sometimes even only D. Definition. A function of the column vectors is a determinant if it satisfies the following three axioms: 1. Viewed as a function of any column A, it is linear and homogeneous, i.e . , ( 3 ) D, (A, + D,( Ak) Dk( cAk) = c ·D,(~ ) (4) 2. Its value is = ~ ) = (\ (A,) + on if the adjacent columns A, and A.k+t are equal. 3. Its value is = 1 if all A, are the unit vectors Uk where 1) Henceforth, 0 will denote the zero element of a field. 1 1 0 0 0 1 ; u, 0 0 0 un 0 1 The question as to whether determinants exist will be left open for the present. But we derive consequences from the axioms: a) If we put c :::: 0 in (4) we get: a determinant is 0 if one of the columns is 0. b) Dk(Ak) Dk(Ak + cA + fr a determinant remains unchanged 15 1 if we add a multiple of one column to an adjacent column. Indeed because of axiom 2. c) Consider the two columns A, and Ak ... l' We may replace them by A, and Ak+l + Ak; subtracting the second from the first we may replace them by - Ak+l and Ak+l + A, adding the first to the second we now have .... Ak+l and A, finally, we factor out -1. We conclude: a determinant changes sign if we interchange two adjacent columns. d) A determinant vanishes if any two of its columns are equal. Indeed, we may bring the two columns side by side after an interchange of adjacent columns and then use axiom 2. In the same way as in b) and c) we may now prove the more general rules: e) Adding a multiple of one column to another does not change the value of the determinant. f) Interchanging any two columns changes the sign of D. 14 g) Let ( v1 , v 2 , . , , v n) be a permutation of the subscripts (1, 2,.... n). If we reanange the columns in D( Av,I Av, .. 2 . , A, ) n until they are back in the natural order, we see that ±. D (A" A 2 , ••• , A n ). D ( Av , Av , ... , A, ) = 1 2 n Here i is a definite sign that does not depend on the special values of the A,. If we substitute Uk for A, we see that D ( U1, , 1 Uv , ... , U 2 Vn ) = ±. 1 and that the sign depends only on the permutation of the unit vectors. Now we replace tionA~ of each vector A, by the following linear combina· A.,, A2 , ••. , An: (6) A; = b1 kA 1 + b2 •A, + . . . + bnkAn. In computing D( A~, A~ , ... , A~) we first apply axiom 1 on A: breaking up the determinant into a sum; then in each term we do the same with A~ and so on. We get (7) D(A;,A;, ... ,A~)= v1 L vl,v2,, '' I •v 2 , vn ,. D(bv 1 Av,bv 2 Av,···'bvnAv) 2 I ,vn J 1 bv 1·b, 2 • •.• n ~ 2 nn 1 2 where each vi runs independently from 1 to n. Should two of the indices vi be equal, then D( Av , A, , ... , Av ) = 0; we need therefore keep 1 n 2 only those terras in which (vi, v , . .. , v0 2 ) is a permutation of ( 1, 2, ... , n). This gives D(A; ,A~, ... ,A~) (8) = D( A1, A2 , " •, A,). ( v1 , ~ , , , , l/n) :_ bv t' bv 1 22 . . . . • bv. 0 n where ( v1 , v2 , ••• , vn) runs through all the permutations of ( 1, 2, ... , n) and where i stands for the sign associated with that permutation. It is impm1ant to remark that we would have anived at the n ·bv D(Av ,A, . . . ,A, ) same formula (8) if our function D satisfied only the first two n 1.5 of our axioms. Many conclusions may be derived from (8). We first assume axiom 3 and specialize the ~ to the unit veetors of (5). This makes ~I == Bk where Bk is the column vector of uk the matrix of the bik' (8) yields now: (9) giving us an explicit formula for determinants and showing that they are uniquely determined by our axioms provided they exist at all. With expression (9) we return to formula (8) and get This is the so-called multiplication theorem for determinants. At the left of (10) we have the determinant of an n-rowed matrix whose ele- ments Cik are given by (11) c. tk = ~" atvbvk' V=l cik is obtained by multiplying the elements of the i .. th row of D(A 1 , A 2 , ••• , An) by those of the k-th column of D(B 1 , 8 2 , .•• , Bn) and adding. Let us now replace D in (8) by a function F(A, satisfies only the first two axioms. Comparing with (9) we find F (A; , A; , ... , A~) ~ F (A 1 , ... , Specializing A, to the unit vectors U k leads to (12) , A,) that F(B 1 ,B 2 ,. with ,B.)=c·D(B 1 ,B 2 ,. c = F(U 1 , U 2 , ... , Un). An )D ( 8 1 , 8 2 , ... , B,). 16 Next we specialize (10) in the following way: If subscript from 1 to n-1 we put A, = uk for II ,;, i is a ce11ain i, i+ 1 A1 = Ui + Ui+l, A1+1 = 0. Then D( A, A, , .. , A, ) = 0 since one column is 0. Thus, D( A~ 1 A~,, .. , A~) = 0; but this determinant differs from that of the elements bJk only in the respect that the i+l-st row has been made equal to the i-tb. We therefore see: A determinant vanishes if two adjacent rows are equal. Each term in (9) is a product where precisely one factor comes from a given row, say, the i-th. This shows that the determinant is linear and homogeneous if s;:onsidered as function of this row. If, finally, we select for eaeh row the corresponding unit vector, th~ terminant is 1 since the matrix is the = s~m~ as that in de- which the col.,. umns are unit vectors. This shows that a determinant satisfies oyr three a:x:ioms if we consider it as function of the row vectors. In view of the uniqueness it follows: A determinant remains unchanged if we transpose the row vectors into coltJil'!ll vectors, that is, if we rotate the matrix about its main diagonal. A determinant vanishes if any two rows are equal. It changes sign if we interchange aqy two rows. It a multiple of on~ rerp~ins unchanged if we add row to another. We shall now prove the existence of determinants. For a 1-rowed mahix a 11 the element a 11 itself is the determinant. Let us assume the existence of (n • 1) • rowed determinants. If we consider the n-rowed mahix (l) we may associate with it certain (n • 1) • rowed determinants in the following way: Let qik be a pmticular element in (1). We 17 cancel the i-th row and k-th column in (1) and take the determinant of the remaining (n .. 1) • rowed matrix. This determinant multiplied by ( -1 )Hk will be called the cofactor of a ik and be denoted by Aik. The distribution of the sign (- l)i+k follows the chessboard pattern, namely, + + + - + + - + - + - + ' ' ' Let i be any number from 1 to n. We consider the following function D of the rnauix (1): D = attAit + ai2Ai2 + · · + (13) [t 8 inAin • is the sum of the products of the i-th row and their cofactors. Consider this D in its dependence on a given column, say, A,. . For v f:- k, A~ depends linearly on A, and a ·v does not depend on it; for v == k, Ark does not depend on A, but a 1k is one element of this column. Thus, axiom 1 is satisfied. Assume next that two adjacent columns A, and Ak+ 1 are equal. For 1J ~ k, k + 1 we have then two equal columns in Ail! so that A, :::: 0. The determinants used in the computation of A 1 hence, A i k:::: -Ai k and k+I Ai k+ 1 whereas a i holds. For the special case A, a;v = 0 for v t are the same but the signs are opposite k :::: a, k+r Thus D :::: 0 and axiom 2 = Uv( v = 1, 2, .. , n) we have i while a,, = 1, A; 1 = 1. Hence, D = 1 and this is axiom 3. This proves both the existence of an n-rowed 18 determinant as well as the truth of formula (13), the so-called development of a determinant according to its i-th row. (13) may be generalized as follows: In our determinant replace the i-th row by the j-th row and develop according to this new row. For i f: j that determinant is 0 and for i = j it is D: { Dforj~i Oforj,ii If we interchange the rows and the columns we get the following fmmula: Dforho=k { Oforh,ik Now let A represent an n-rowed and B an m-rowed square matrix. By \ A \, \ B \ we mean their determinants. Let C be a mahix of n rows and m columns and form the square matrix of n + m rows (16) where 0 stands for a zero matrix with m rows and n columns. If we consider the determinant of the matrix (16) as a function ofthecolumns of A only, it satisfies obviously the first two of our axioms. Because of (12) its value is c . A where c is the determinant of (16) after substituting unit vectors for the columns of A. This c still depends on B and considered as function of the rows of B satisfies the first two axioms. Therefore the determinant of (16) is d · A . B where d is the special case of the determinant of (16) with unit vectors for the columns of A as well as of B. Subtracting multiples of the columns of A from C we can replace C by 0. This shows d = 1 and hence the formula 19 (17) l AO CBI ~ I AI. I Bi In a similar fashion we could have shown (18) A 01 \AI \BI C B The formulas (17), (18) are special cases of a general theorem by Lagrange that can be derived from them. We refer the reader to any textbook on determinants since in most applications (17) and (18) are sufficient. We now investigate what it means for a matrix if its determinant is zero. We can easily establish the following facts: a) If A, A, . , . , An are linearly dependent, then D (A,. A, ... , A,) = 0. Indeed one of the vectors, say A, is then a linear combination of the other columns; subtracting this linear combination from the column A, reduces it to 0 and so D = 0. b) If any vector B can be expressed as linear combination of A, A, ... , A, then D (A,. A 2 , •.• , (10) we may select the values for Al A,) ~ 0. Returning to (6) and bik in such a fashion that every .~ U1 • For this choice the left side in (10) is 1 and hence D (A,. A 2 , ••. , A,) on the right side ~ 0. c) Let A, A, ... , A, be linearly independent and B any other vector. If we go back to the components in the equation A 1 X 1 + A 2x 2 + . . . + Anxn+ By= 0 we obtain n linear homogeneous equations in then + 1 unknowns x 1, x 2 , . . . , xn, y. Consequently, there is a non-trivial solution. y must be f=- 0 or else the A 1, A 2 , ... , ~ would be linearly dependent. But then we can compute B out of this equation as a linear combination of A, A, . . An· 20 Combining these results we obtain: A determinant vanishes if and only if the column vectors (or the row vectors) are linearly dependent. Another way of expressing this result is: The set of n linear homogeneous equations ( i = 1, 2, ... , n) in n unknowns has a non-trivial solution if and only if the determinant of the coefficients is zero. Another result that can be deduced is: If A I' A 2 , ... , A, are given, then their linear combinations can represent any other vector B if and only if D (A , An ) , A, 1 ~ o. Or: The set of linear equations (i = 1, 2, ... , n) has a solution for arbitrary values of the bi if and only if the determinant of aik is f 0. In that case the solution is unique. We finally express the solution of (19) by means of determinants if the determinant D of the aik is -f 0. We multiply for a given k the i-th equation with Aik and add the equations. (15) gives (20) D· xk = A1kb 1 c A 2 kb 2 c Ankbn (k = 1,2, ... ,n) and this gives xk. The right side in (12) may also be written as the determinant obtained from D by replacing the k-th column by b, b 2' bn. The rule thus obtained is known as Cramer's rule. 21 II FIELD THEORY A. Extension Fields. If E is a field and F a subset of E which, under the operations of addition and multiplication in E, itself forms a field, that is, if F is a subfield of E, then we shall call E an extension of F. The relation of being an extension of F will be briefly designated by F C E. If a, (3, y, ... are elements of E, then by F(a, (3, y, ... ) we shall mean the set of elements in E which can be expressed as quotients of polynomials in a, F (a, f3, f3, y, ... y, .. with coefficients in F. It is clear that ) is a field and is the smallest extension of F which con- tains the elements a, (3, y, .. We shall call F(a, fl, y, . . . ) the field obtained after the adjunction of the elements a, {3, y, . .. to F, or the field generated out of F by the elements a, {3, y, .... In the sequel all fields will be assumed commutative. If F c E, then ignoring the operation of multiplication defined between the elements of E, we may consider E as a vector space over F. By the degree of E over F, written (E/F), we shall mean the dimension of the vector space E over F. If (E/F) is finite, E will be called a finite extension. THEOREM 6. If F, B, E are three fields such that F c B c E, then (E/F) = (8/F) (E/8) . Let A1 , A 2 ,. , • , Ar be elements of E which are linearly independent with respect to Band let C 1, C 2 , ... , C s be elements 22 of B which are independent with respect to F. Then the products Ci Ai where i = 1, 2, ... , s and j = 1, 2, ... , r are elements of E which are For if ~ a 11 C 1A; = 0, then independent with respect to F. :?( '·' 2:a 1J C. ) A,. is a linear combination of the A, with coefficients in B i l ' and because the A i were independent with respect to B we have l:a .. C 1 = 0 for each j. The independence of the C 1 with respect to F i lJ then requires that each ail = 0. Since there are r . s elements CiAi we have shown that for each r 2_ ( E/B) and s ::_ ( B/F) the degree ( E/F ) ;: r . s. Therefore, ( E/F) ::_ ( 8/F) ( E/8 ). If one of the latter numbers is infinite, the theorem follows. If both ( E/8) and ( B/F) are finite, say r and s respectively, we may suppose that the Aj and the C 1 are generating systems of E and B respectively, and we show that the set of products Ci A1 is a generating system of E over F. Each A f E can be expressed linearly in terms of the A. with coefficients in B. Thus, J A = lBJ A .. Moreover, each B. being an element of B can be ex' J pressed linearly with coefficients in F in terms of the Ci' i.e., Bj = 2:a 1j CFj = 1,2, . . . ,r. Thus, A= ~aij CiAj and the C 1 ~ form an independent generating system of E over F. ~orollary._lf F C F 1 C F2 C . . . C Fn, then (Fn/F) c (F/F)·(F2 /F,) . . . (Fn/Fn_ 1 ). B. Polvnomials. An expression of the form a 0 xn +a 1 xn-t+ ... + a 0 is called a polynomia~ in F of degree n if the coefficients 23 a 0 , ... , a,., are elements of the field F and a 0 f: 0. Multiplication and addition of polynomials are performed in the usual way 0 . A polynomial in F is called reducible in F if it is equal to the product of two polynomials in F each of degree at least one. Polynomials which are not reducible in F are called ineducible in F. If f (x ) = g(x) . h (x ) is a relation which holds between the polynomials f (x ), g (x ), h (x ) in a field F, then we shall say that g (x ) divides f (x ) in F, or that g ( x ) is a factor off ( x ). It is readily seen that the degree of f(x) is equal to the sum of the degrees of g (x ) and h (x ), so that if neither g ( x ) nor h ( x ) is a constant then each has a degree less than f(x). It follows from this that by a finite number of factorizations a polynomial can always be expressed as a product of ineducible polynomials in a field F. For any two polynomials f (x ) and g (x ) the division algorithm holds, i.e., f(x) = q(x)·g(x) + r(x) where q(x) and r(x) are unique polynomials in F and the degree of r (x ) is less than that of g(x). This may be shown by the same argument as the reader met in elementary algebra in the case of the field of real or complex numbers. We also see that r(x) is the uniquely determined polynomial of a degree less than that of g (x ) such that f(x) - r (x ) is divisible by g (x ). We shall call r (x ) the remainder off (x ). 1) If we speak of the set of all polynomials of degree lower than n, we shall agree to include the polynomial 0 in this set, though it has no degree in the proper sense. 24 Also, in the usual way, it may be shown that if a is a root of the polynomial f (x ) in F than x - a is a factor off (x ), and as a con- sequence of this that a polynomial in a field cannot have more roots in the field than its degree. Lemm<!_. If f(x) is an irreducible polynomial of degree n in F, then ther~ do not exist two polynomials each of degree less than n in F whose _product is divisible by f(x). Let us suppose to the contrary that g(x) and h(x) are polynomials of degree less than n whose product is divisible by f(x). Among all polynomials occmring in such pairs we may suppose g(x) has the smallest degree. Then since f(x) is a factor of g(x) . h (x ) there is a polynomial k(x) such that k(x)·f(x) = g(x)·h(x) By the division algorithm, f(x) = q(x).g(x) + r(x) where the degree of r (x ) is less than that of g(x) and r (x ) ~ 0 since f(x) was assumed ineducible. Multiplying f(x) = q(x).g(x) + r(x) by h (x ) and transposing, we have r (X) · h (X) = f (X)· h (X )-q (X ) · g (X) · h (X) - f (X)· h (X )-q (X )· k (X) ·f (X) from which it follows that r(x) . h (x ) is divisible by f (x ). Since r (x ) has a smaller degree than g(x), this last is in contradiction to the choice of g (x )~ from which the lemma follows. As we saw, many of the theorems of elementary algebra hold in any field F. However, the so-called Fundamental Theorem of Algebra, at least in its customary form, does not hold. It will be replaced by a theorem due to Kronecker 25 which guarantees for a given polynomial in F the existence of an extension field in which the polynomial has a root. We shall also show that, in a given field, a polynomial can not only be factored into irredllcibl~ factors, but that this factorization is unique up to a constant factor. The uniqueness depends on the theorem of Kronecker. C. Algebraic Elements. Let F be a field and E an extension field of F. If a is an element of E we may ask whether there are polynomials with coefficients in F which have a as root. a is tkere are ~alled algebraic with respect to F if such polynomials. Now let a be algebraic and select among all polynomials in F which have a as root one, f(x), of lowest degree. We mpy assume that the highest coefficient of f(x) i~ 1. We con- tend that this f(x) ~~ uniquely determined, that it is irreducible and that each polynomial in F with the r1>0t a is divisible by f c~ ). If. in- deed, g ( ~ ) i& a pQ!ynomial in F with g(a) = 0, we may divide g(x) = f(x)q(x) t r(x) where r(x) ha~ a degree sm~ller than that of f(x). Substituting x = a we get r ( n l i<lenti~ally 0 since otherwis@ r ( ~ ) lo\Ver degree thap f (x ), the uniqueness of f (x So g ( x ) woul~ j~ = 0. NPW r(x) has to be h§Ve the root divisible by f ( • ), a and be of Th!~ also shows ), Iff ( N ) were not irreducible, one of the factors Would have to vanish for x =a contraqict\ng again the choice off ( y ), We consider now the subset Eo of the following 0 ofE: element~ 26 (J = g(a) = Co+ Cla + c2a2 + ... + cn-Ian-1 where g(x) is a polynomial in F of degree less than n (n being the degree of f(x)). This set tf, is closed under addition and multiplication. The latter may be verified as follows: If g (x ) and h (x ) are two polynomials of degree less than n we put g(x)h(x) = q(x)f(x) + r(x) and hence g(a)h(a) = r(a). Finally we see that the constants c , c , .. , c are uniquely deter1 ' n- 1 mined by the element 0. Indeed two expressions for the same f) would lead after subtracting to an equation for a of lower degree than n. We remark that the internal structure of the set E 0 does not de- pend on the nature of a but only on the ineducible f (x ). The knowledge of this polynomial enables us to perform the operations of addition and multiplication in our set E 0 • We shall see very soon that E 0 is a field; in fact, Eo is nothing but the field F(a). As soon as this is shown we have at once the degree, ( F (a) /F), determined as n, since the space F(a) is generated by the linearly independent 1, a, a2, . .. , an-I, We shall now try to imitate the set E 0 without having an extension field E and an element a at our disposal. We shall assume only an ineducible polynomial as given. We select a symbol I; and let E 1 be the set of all formal polynomials g(f;) =co+ elf;+ .. + cn_,f;n-1 of a degree lower than n. This set forms a group under addition. We now introduce besides the ordinary multiplication 27 a new kind of multiplication of two elements g (,f) and h (,f) of E 1 denoted by g (,f) x h (<fl. It is defined as the remainder r (.f) of the ordinary product g (,f) h (,f) under division by f (f). We first remark that any product of m terms g 1 ( ,f), g 2 ( f),, .. 1 gm( mainder of the ordinary product g 1( ~) g ,( f) is again there- 0 ... g'"( ~). This is true by definition for m = 2 and follows for every m by induction if we just prove the easy lemma: The remainder of the product of two remainders (of two polynomials) is the remainder of the product of these two polynomials. This fact shows that our new product is associative and commutative and also that the new product g ( 1 f) x g 2( ,f) x , , , x g mC<f) will coincide with the old product g 1( ,f) g 2( ~) •. , gm( ,f) if the latter does not exceed n in degree. The distributive law for our multiplication is readily verified. The set E 1 contains our field F and our multiplication in E1 has for F the meaning of the old multiplication. One of the polynomials of E 1 is f Multiplying it i-times with itself, clearly will just lead to as long as i < n. <f' For i : : : n this is not any more the case since it leads to the remainder of the polynomial I;. This remainder is = - a n- t n-1 - an- ~n-2 - ' ' ' - ao. We now give up our old multiplication altogether and keep only the new one; we also change notation, using the point (or juxtaposition) as symbol for the new multiplication. Computing in this sense Co + c CtS + /:2 C2t.:, +' ·+ en-! Cn-tS will readily lead to this element, since all the degrees 28 involved are below n. But Transposing we see that f ( el = 0. We thus have constructed a set E 1 and an addition and multipli- cation in E 1 that already satisfies most of the field axioms. E 1 contains F as subfield and satisfies the equation f ( ~) = 0. We next have to e show: If g ( f) i 0 and h ( ~)are given elements of E 1, there is an element X( f)= •. + x 1 ~ x._,f•·' + ... + in E l such that &<e). xcel = h(eJ. To prove it we consider the coefficients x. of X ( ~) as unknowns and ' . compute nevertheless the product on the left side, always reducing higher powers of ~ to lower ones. The result is an expression L o + L 1 ~ + . . + Ln-t en-t where each L 1 is a linear combination of of the Xr with coefficients in F. This expression is to be equal to h ( {); this leads to the n equations with n unknowns: Lo = bo, Ll = bl' · · · 1 Ln-l = bn-1 where the b1 are the coefficients of h (f). This system will be soluble if the conesponding homogeneous equations L. = 0, L, = 0, ' .. have I L •. , 0 only the trivial solution. The homogeneous probleiP would occur if we should ask for the set of elements X(Q) satisfying g ( 0 . X ( el = 0. Going back for a moment to the old multiplication this would mean that the ordinary product g ( 0 X ( eJ has the remainder 0, and is 29 therefore divisible by f(t). According to the lemma, page 24, this is only possible for X (f) = 0. Therefore E 1 is a field. Assume now that we have also our old extension E with a root a of f(x), leading to the set E,. We see that E 0 has in a certain sense the same structure as E 1 if we map the element g (,;)of E 1 onto the element g(a) of E0 • This mapping will have the prope11y that the image of a sum of elements is the sum of the images, and the image of a product is the product of the images. Let us therefore define: A mapping a of one field onto another which is one to one in both directions such that a(a+f3) = a(a) + a({3) and a(a·{3)= a(a)· a({3) is called an isomorphism. If the fields in question are not distinct - i.e., are both the same field - the isomorphism is called an automorphism. Two fields for which there exists an isomorphism mapping one on another are called isomorphic. If not every element of the image field is the image under a of an element in the first field, then o is called an isomorphism of the first field into the second. Under each isomorphism it is clear that a(O) = 0 and a( 1) = 1. We see that Eo is also a field and that it is isomorphic to E,. We now mention a few theorems that follow from our discussion: THEOREM 7. (Kronecker). If f (x ) is a polynomial in a field F, there exists an extension E of F in which f(x) has a root. 30 Proof: Construct an extension field in which an irreducible factor off ( x ) has a root. THEOREM 8. Let a be an isomorphism mapping a field F on a field F' Let f (x ) be an irreducible polynomial in F and f responding polynomial in F 1 • If E = F ( f3) and E a can = F 1 ( = 0 in E and (x ) the cor- f~') are exten- f 1 be extended to an isomorphism between E and E 1 sions ofF and F', respectively, where f(f3) then 1 1 ( f3 ') = 0 in E' Proof: E and E' are both isomorphic to E 0 • D. Splitting Fields. If F, B and E are three fields such that F C B C E, then we shall refer to B as an intermediate field. If E is an extension of a field F in which a polynomial p(x) in F can be factored into linear factors, and if p(x) can not be so factored in any intermediate field, then we call E a splitting field for p(x). Thus, if E is a splitting field of p(x), the roots of p(x) generate E. A splitting field is of finite degree since it is constructed by a finite number of adjunctions of algebraic elements, each defining an extension field of finite degree. Because of the corollary on page 22, the total degree is finite. THEOREM 9. If p(x) is a polynomial in a field F, there exists a splitting field E of p(x). We factor p (x ) in F into irreducible factors f,(x) f,( x) = p(x). If each of these is of the first degree then F itself is the required splitting field. Suppose then that f 1 ( x) is of degree higher than the first. By 31 Theorem 7 there is an extension F 1 ofF in which f Factor each of the factors f 1 ( ( 1 x ) has a root. x), ... , fr ( x ) into irreducible factors in F 1 and proceed as before. We finally arrive at a field in which p (x) can be split into linear factors. The field generated out of F by the roots of p(x) is the required splitting field. The following theorem asserts that up to isomorphisms, the splitting field of a polynomial is unique. THEOREM 10. Let the field F' , ~~ith a be an isomorphism mapping the field F on Let p (x ) be a polynomial in F and p 1 (x ) the polynomial coefficients corresponding to those of p (x ) under let E be a splitting field of p(x) and E Under these conditions the isomorphism isomorphism between E and E If f(x) is an irreducible 1 1 1) ( a can in which f(a) = 0. Then (a-a be extended to an • factor of p(x) in F, then E contains a x-a 2 ). f(x) as a polynomial in E and Finally, a splitting field of p 1 (x). root of f( x ). For let p (x )=(x-a ~ (x-a, ) p(x) in E. Then ( x-a a. •( (x-a ,) be the splitting of x-a ,) = f(x) g(x). We consider construct the extension field B = E(a) 1 )· (a-a 2 )· .•. ·(a-a,)= f(a)· g(a) = 0 and a-a1 being elements of the field B can have a product equal to 0 only if for one of the factors, say the first, we have a-a 1 = 0. Thus, a = a 1 , and a 1 is aroot of f(x). Now in case all roots of p(x) are in F, then E = F and p(x) can be split in F. This factored form has an image in F splitting, of p 1 (x), since the isomorphism a I which is a preserves all operations of addition and multiplication in the process of multiplying out the factors of p(x) and collecting to get the original form. Since p ' (x) can be split in F' , we must have F 1 :::: E 1 • In this case, a itself is the required extension and the theorem is proved if all roots of p(x) ar~ in F. We proceed by complete induction. Let us suppose the theorem proved for all cases in which the number of roots of p(x) outside of F is less than n > 1, and suppose that p (x ) is a polynomial having n roots outside of F. We factor p (x ) into ineducible factors in F; p(x) = f,(x) f (x) .. f,(x). Not all of these factors can be of 2 degree 1, since otherwise p (x ) would split in F, contrary to assumption. Hence, we may suppose the degree of f 1( x) to be r > 1. Let f\(X).f'2(x) . . . fO:,(x) = p'(x) be the factorization of p'(x) into the polynomials comespondng to f is ineducible in F 1 , ( 1 x ) , ... , f m ( x ) under a. for a factorization of f; (x) in F 1 £; (x ) would induce 1) under a-1 a factorization of f,(x), which was however taken to be irreducible. By Theorem 8, the isomorphism a can be extended to an isomor- phism a 1, between the fields F (a) and F ' (a 1 ). Since F C F(a), p(x) is a polynomial in F (a) and E is a splitting field for p(x) in F(a). Similarly for p ' (x). There are now less than n roots of p (x ) outside the new ground field F (a). Hence by our inductive assumption a 1 can be extended from an isomorphism between F (a) and F 1 (a ' ) to an isomorphism a 2 between E and E Since a1 is an extension of a, and a2 an extension of a 2 is an extension of a and the theorem follows. t) See page 38 for the definition of a a 1, we 1 • conclude 33 Corollary. If p ( x) is a polynomial in a field F, then any two splitting fields for p ( x ) are isomorphic. This follows from Theorem 10 if we take F = F ' and a to be the identity mapping, i.e., a(x) = x. As a consequence of this corollary we see that we are justified in using the expression "the splitting field of p ( x )" since any two differ only by an isomorphism. Thus, if p (x ) has repeated roots in one splitting field, so also in any other splitting field it will have repeated roots. The statement up ( x) has repeated roots" will be significant without reference to a particular splitting field. E. Unique Decomposition of Polynomials into Irreducible Factors. THEOREM 11. If p( x) is a polynomial in a field F, and if p(x) = p 1(x)·p 2 (x)· .. , ·p,(x) = q 1(x)·q 2(x)· .... q (x) are two factorizations of p ( x) into irreducible polynomials each of de~ree at least one, then r = s and after a suitable change in the order in which the q's are written, p,(x) ~ c,q.(x), i = 1,2,., ,r, and c,' F. Let F (a) be an extension of F in which p 1( a) ~ 0. We may suppose the leading coefficients of the p,( x) and the q1(X) to be 1, for, by factoring out all leading coefficients and combining, the constant multipliE~r on each side of the equation must be the leading coefficient of p ( x ) and hence can be divided out of both sides of the equation. Since 0 = p 1(a)· p 2 ( a) · , , , · P, (a) = p (a) ~ q 1( a) · . , , · q ,(a) and since a product of elements of F (a) can be 0 only if one of these is 0, it follows that one of the q,( a), say q 1( a), is 0. This gives (see page 25) p 1(>) = q 1(x ). Thus p 1(x). p 2(x) ..... p,(x) = p 1(x)·q 2 (x)·,, ·q,(x) or H p 1(x) ·[p 2(x) · .. ·p,(x)- q 2 (x) · .. ·q,(x)] = 0. Since the product of two polynomials is 0 only if one of the two is the 0 polynomial, it follows that the polynomial within the brackets is 0 so that · Pr ( x) = q 2( X)· ... · q 5( X). If we repeat the above argument p,(x) r times we obtain pi ( x) = q i (X). i = 1, 2,. . r. Since the remaining q's must have a product 1, it follows that r-: s. F. ~roup C~l!_racter~. If G is a multiplicative group, F a field and a a homomorphism mapping G into F, then a is called a character of G in F. By homomorphism is meant a mapping a such that for a, f3 any two elements of G, a( a )·a({3) = a(a·{J) and a(a) ~ Oforanya. (If o(a) = 0 for one element a, then o(x) = 0 for each x ( G, since a( ay) = a( a:)· o(y) ""0 and ay takes all values in G when y assumes all values in G). The characters a elements a 1 , a, . 1 , a 2 ,. . , a nare called dependent if there exist . , a, not all zero in F such that . + ana n( x) ::: 0 for each x ( G. Such a de- pendence relation is called non-trivial. If the characters are not dependent they are called independent. THEOREM 12. If G is a group and a 1' a 2, •.. , an are n mutually distinct characters of G in a field F, then a 11 a 2,. . an are independent. One character cannot be dependent, since a a1 1a 1( x) ::: 0 implies 0 due to the assumption that a (x) ~ 0. Suppose n 1 > 1. 35 We make the inductive assumption that no set of less than n distinct characters is dependent. Suppose now that a 0 a 0 ( x) :::: 0 is a non-trivial dependence t between the a's. None of the elements ai is zero, else we should have a dependence between less than n characters contrary to our induetive assumption. Since o a in Gi such that a1 (a) ~ 1 and an are distinct, there exists an element an( a). Multiply the relation between the a's b y n a ·We obtain a relation (*) b 1 a 1 (x)t +b0 _ 1 u 0 _ 1 (x)to,(x) Replace in this relation x by ax. We have 0, Subtracting the latter from ( *) we have (**) [b 1 an Car'b 1 a 1 (a)]a 1 (x)'. + C 0 _ 1 0 0 _ 1 (x) o . = The coefficient of a 1 (x ) in this relation is not 0, otherwise we should have b 1 = o-0 (a)" 1 b 1 a 1 (a), so that an (a)b 1 = b 1 a 1 (a)= u 1 (a)b 1 and since b 1 ~ 0, we get a,( a) = a 1 (a) contrary to the choice of a. Thus, (* *) is a non-trivial dependence between a l' a2, , • an- 1 which is contrary to our inductive assumption. Corollary. If E and E 1 are two fields, and a 1 , a2 mutually distinct isomorphisms mapping E into E 1 , an are n , .. , then u 1 , , an are independent. (Where ttindependent" again means there exists no non-trivial dependence a every x f 1a 1 (x ) + . + anan (x ) == 0 which holds for E). This follows from Theorem 12, since E without the 0 is a group 36 a's defined in this group are mutually distinct characters. and the If a 1 , o2 1 then each ••• , an are isomorphisms of a field E into a field E' element a of E such that is called a fixed point of E under a 1 (a) = o,(a) = a1 , on . This name is 02 , . chosen because in the case where the a's are automorphisms and a1 is the identity, i.e. 1 a1 (x) = x, we have ai (x) = x for a fixed point. Lemma. The set of fixed points of E is a subfield of E. We shall call this subfield the fixed field. For if a and b are fixed points, then a1 (a +b)= a 1 (a) + a1 (b) =";(a)+ Oj (b)="; (a+ b) and a,(a·b) = a 1 (a)·a,(b) = "; (a)·a;(b) =a; (a·b). Finally from = ",ca- 1 ) a 1 (a) = a, (a) we have (a,(a))- 1 = (a 1 (alr' =a; (a-'). Thus, the sum and product of two fixed points is a fixed point, and the inverse of a fixed point is a fixed point. Clearly, the negative of a fixed point is a fixed point. THEOREM 13. If a 1'. of a field E into a field E 1 , . 1 an are n mutually distinct isomorphisms and if F is the fixed field of E, then (E/F) >_n. Suppose to the contrary that ( E/F) = r < n. We shall show that we are led to a contradiction. Let w 1, w 21 . . . , wr be a generating sys- tern of E over F. In the homogeneous linear equations 37 u 1 (w 1 )x 1 + a 2 (w 1 )x 2 + .. + a 0 (w 1 )xn = 0 u 1 (w 2 )x 1 + o- 2 (w 2 )x 2 + .. + an(w,)xn = 0 a 1 (cur)x 1 + a 2 (wr)x 2 + ... + a 0 (W 1 )X = 0 0 there are more unknowns than equations so that there exists a nontrivial solution which, we may suppose, x any element a in Ewe can find a 1 , a 2 , 1, x •.. , denotes. For 2 , . . . , X0 a, in F such that a ::. a 1t:v 1 + ... + a/ur. We multiply the first equation by the second by a 1 (a 2 ), and so on. Using that a 1 a 1 (aJ) = a 1 (a 1 )and also that a (a) 1 we obtain a 1 (arwr)x 1 + ... + 0 0 a;(w 1) f. F, a 1 (a 1 ), hence that = a (a 1w 1 ), 1 (arwr)x0 = 0. Adding these last equations and using ai(a 1 U) 1) 1- a 1 (a 2 (t.~ 2 ) + ... + aJarcu) = a 1 (a 1w 1 + ... + arwr) = ai(a) we obtain a 1 (a)x 1 + a 2 (a)x 2 + . . . + un(a)xn = 0. This, however, is a non-trivial dependence relation between a 11 a 2 , . which cannot exist according to the corollary of Theorem 12. Corollary. If a 1, a 2, ••• , an are automorphisms of the field E, and F is the fixed field, then ( E/F) > n. If F is a subfield of the field E, and shall say that a leaves F fixed if for a an automorphism of E, we each element a of F, o(a) = a. a0 38 If a and r are two automorphisms of E, then the mapping o-( r( x)) written briefly ar is an automorphism, as the reader may readily verify. [E.g., <n(x y) ~ a(r(x · y)) We shall call (a(x) ~ = a(r(x)· r(y)) = a(r (x)) · a(r(y))]. ar the product of a and r. If a is an automorphism y), then we shall call a· 1 the mapping of y into x, i.e.,a· 1(y) ~ the inverse of a. The reader may readily verify that a·' is an automor· phism. The automorphism I (x) ~ x shall be called the unit automorphism. Le~. If E is an extension field of F, the set G of automorphisms which leave F fixed is a group. The product of two automorphisms which leave F fixed clearly leaves F fixed. Also, the inverse of any automorphism in G is in G. The reader will observe that G, the set of automorphisms which leave F fixed, does not necessarily have F as its fixed field. It may be that certain elements in E which do not belong to F are left fixed by every automorphism which leaves F fixed. Thus, the fixed field of G may be larger than F. G. Applications and Examples to Theorem 13. Theorem 13 is very powerful as the following examples show: 1) Let k be a field and consider the field E = k (x ) of all rational functions of the variable x. If we map each of the functions f (x) of E onto f(!) we obviously obtain an automorphism of E. Let US X consider the following six automorphisms where f (x ) is mapped onto 1 1 ) and f f (x) (identity), f ( 1-x), f ( ·), f (1·! ), f CX X 1-X C-"-l and call X-1 F the x 39 fixed point field. F consists of all rational functions satisfYing (1) It 1 = f(l-x) 1 = f(r:xl 1 = f( x-1 X ) f(x) = f(l-x) = f(x) . suffices to check the first two equalities, the others being conse- quences. The function I= I(x)- (x'- X+ 1 )' x 2 (x-1) 2 belongs to F as is readily seen. Hence, the field S = k (1) of all (2) rational functions of 1 will belong to F. We contend: F = Sand ( E/F) = 6. lndeed, from Theorem 13 we obtain (E/F) 2 6. Since S C Fit suffices to prove (Ef S) < 6. Now E = Sx). It is thus sufficient to find some 6-th degree equation with coefficients in S satisfied by x. The following one is obviously satisfied; (x 2 -x+1) 3 -1-x 2 (x-1) 2 = 0. The reader will find the study of these fields a profitable exer- cise. At a later occasion he will be able to derive all intermediate fields. 2) Let k be a field and E = k(x 1 ,x 2 , rational functions of n variables x 1 , x 2 , ... , X ... • 0 ,xn)the field of all If ( v 1 , v 2 , permutation of { 1, 2, ... , n) we replace in each function f (x of E the variable X1 by X , X vl 2 by X ~ ,. . ... , 1 , vn ) is a x 2 , . . . , xn) , X by XV . The mapping of E n n onto itself obtained in this way is obviously an automorphism and we may construct n ! automo:rphisms in this fashion (including the identity). Let F be the fixed point field, that is, the set of all so-called "symmetric functions." Theorem 13 shows that ( E/F) >_ n ! . Let us introduce the polynomial: (3) f(t) = (t-x 1 )(t-x 2 ) ... (t-x,) tn + a 1 t n-1 + . + an 40 wherea 1 = ... (x 1 + x 2 + ... + x,); a 2 (x 1 x 2 + x 1 x3 + ... + :: + xn-lxn) and more generally ai is ( • 1)' times the sum of all products of i differerentvariables of the set x 1 , x 2 , ... , xn.Thefunctionsa, a 2 , ... , an are called the elementary symmetric functions and the field S = k ( a1 , a2 , .•• , an ) of all rational functions of a,, a 2 , an is ... , obviously a part of F. Should we suceed in proving ( Ef S ) < n ! we would have shown S = F and ( E/F) = n ! . We construct to this effect the following tower of fields: S = Sn C Sn-t C S 0 _ 2 ... C S 2 C S1 = E by the definition (4) sn = S; s± = S(xl+t ,xi+2····,xn) = s±•t (xt+t ). It would be sufficient to prove ( S 1 _1 /S± ) <_ i or that the generator x 1 for Si-l out of ~ satisfies an equation of degree i with coefficients inS1 . Such an equation is easily constructed. Put f(t) FH! (t) (5) F, ( t) = (t-xl+l )(t-x,+2 ) ... (t-xn) = (t-x,+l) and F n ( t) = f ( t ). Perfonning the division we see that F 1 (t ) is a polynomial in t of degree i whose highest coefficient is 1 and whose coefficients are polynomials in the variables a1 , a2, ... , a,, and xi+l, xi+2 in these expressions. Now Now let g(x 1 , x 2 , Since F1 ( 'S ) = , . . . , xn. Only integers enter as coefficients xi is obviously a root of F1 (t ) ••• , xn) be a polynomial in x 1 , x 2 , 0 is of first degree in 'S , we polynomial of the~ and of x 2 , x 3 , . . . , xn. in g( x 1 , x 2 , ) = ••• , xn ). Since F2 ( x 2 can express = 0. •.. , xn. 'S as a We introduce this expression 0 we can express x~ or higher 41 powers as polynomials in x 3 , ... , xn and the a 1 • Since F 3 x3) = 0 ( we can express xj and higher powers as polynomials of x 4 , x 5 ,, .. , xn and the ai. Introducing these expressions in g( x 1 , x 2 , •.. , xn) we see that we can express it as a polynomial in the the degree in xi is below i. SJ g( x 1 , x 2 , ~ ... , X and the ) 0 ~ such that is a linear combination of the following n ! terms: vn (6) ... X 0 where each vi :S_ i - 1. The coefficients of these terms are polynomials in the ai . Since the expressions (6) are linearly independent in S (this is our previous result), the expression is unique. This is a generalization of the theorem of symmetric ftu1ctions in its usual form. The latter says that a symmetric polynomial can be written. as a polynomial in a 1 , a, ... , a,. Indeed, if g(x 1 , ... ,x 0 ) is symmetric we have already an expression as linear combination of the terms (6) where only the term 1 conesponding to v1 = has a coefficient -f 0 inS, namely, g( x 1 , is a polynomial in a, a 2 , •.. , ... , l.-' 2 = ... = vn = 0 xn ). So g( x 1 , x 2 , •.. , xn) a,. Hut our theorem gives an expression of any polynomial, symmetric or not. H. Nmmal Extensions. An extension field E of a field F is called a normal extension if the group G of automorphisms of E which leave F fixed has F for its fixed field, and ( E/F) is finite. Although the result in Theorem 13 cannot be sharpened in general, 42 there is one case in which the equality sign will always occur, namely, in the case in which a1 , a2 , . . . , an is a set of automo:rphisms which form a group. We prove THEOREM 14. If a 1 , u2 , •. . , an is a group of automo:rphisms of a _field E ru:d ifF is the fixed field of u 1 Ifa 1 ,a 2 , ••• ,a, ... , a., then (E/F) = n. ,a0 is a group, then the identity occurs, say, (] 1 =I. The fixed field consists of those elements x which are not moved by any of the a's, i.e., a i ( x) = x, i ::::: 1, 2, ... n. Suppose that ( E/F ) > n. Then there exist n + 1 elements a 1 , a 2 , . . . , an+l of E which are linearly independent with respect to F. By Theorem 1, there exists a non-trivial solution in E to the system of equations 0 xl al (al ) + x2 al (a2 ) + ' , ' + xn+l at (an+l ) -:: Xl a2 (a I ) + x2 a2 ( a2 + ... + ) 0 xn+I a2 ( an+l ) = (' ) x1 a. ( a 1 ) + x 2 an ( a2 ) + , , , + xn+l "" (an+! ) = 0 We note that the solution cannot lie in F, otherwise, since a is the 1 identity, the first equation would be a dependence between a,, Among all non-trivial solutions x 1 x ' 2' ... , x n+l . . . , a n+l we choose one which has the least number of elements different from 0. We may suppose this solution to be a 1 , a 2 , ... , a, 0, ... , 0, where the first r terms are different from 0. Moreover, r implies a 1 = 0 since a 1 (a, ) f. 1 because a, a 1 (a 1 ) = 0 = a 1 -/:. 0. Also, we may suppose ar = 1, since if we multiply the given solution by a; 1 we obtain a new solution in which the r- th term is 1. Thus, we have 43 (*) a 1a,(a 1 ) + a2 a1(a 2 ) + .. , + a,_ 1a,(a,_ 1 ) + a1 (a,) fori= 1, 2, ... , n. these, say a 1 , Since a 1 , . . . , ar-l ~ 0 cannot all belong to F, one of is in E but not in F. There is an automorphism ak for which ak( a,) f. a 1 . If we use the fact that a 1 , a 2 , . .. , an form a group, we see 17k·a 1 ,ak·a 2, ... ,ak·anisapermutationofa 1 ,u 2, . . . ,an. Applyjng ak to the expressions in ( *) we obtain ak(a 1 ).akaj(a 1 ) + ,,, + ok(ar_ 1 )·akaj(ar-I) for j = l, 2, , .. , n, so that from and if we subtract ( akaj + akaj(ar) = 0 =a1 * * ) from ( * ) we have [a 1 - ak(a 1 )l. aJa 1 ) + ... + la,_ 1 - ak(a,_ 1 )]a,(a,_1 ): 0 which is a non- trivial solution to the system C ) having fewer than elements different from 0, contrary to the choice of r. Corollary 1. If F is the fixed field for the finite group G, then each automorphism a that leaves F fixed must belong to G. ( E/F) = order of G = n. Assume there is a a not in G. Then F would remain fixed under the n + 1 elements consisting of (] and the elements of G, thus contradicting the corollary to Theorem 13. Corollary 2. There are no two finite groups G1 and G 2 with the same fixed field. This follows immediately from Corollary 1. If f(x) is a polynomial in F, then f(x) is called separable if its irreducible factors do not have repeated roots. If E is an extension of 44 the field F, the element a of E is called separable if it is root of a separable polynomial f(x) in F, and E is called a separable extension if each element of E is separable. THEOREM 15. E is a normal extension of F if and only if E is !he spli!ting field of a separable polynomial p(x) in F. Sufficiency. Under the assumption that E splits p (x) we prove that E is a normal extension ofF. If all roots of p(x) are in F, then our proposition is trivial, since then E = F and only the unit automorphism leaves F fixed. Let us suppose p(x) has n > 1 roots in E but not in F. We make the inductive assumption that for all pairs of fields with fewer than n roots of p(x) outside of F our proposition holds. Let p(x) = p 1 ( x). p 2 ( x). . . . . p,( x) be a factorization of p(x) into irreducible factors. We may suppose one of these to have a degree greater than one, for otherwise p(x) would split in F. Suppose deg p,(x) = s > 1. Let a be a root of p,(x). Then (F(a, )/F) = deg p,(x) 1 If we consider F (a 1 ) as the new ground field, fewer roots of p ( x) than n are outside. From the fact that p(x) lies in F(a, ) and E is a splitting field of p(x) over F( a 1 ), it follows by our inductive assumption that E is a normal extension of F (a, ). Thus, each element in E which is not in F(a, ) is moved by at least one automorphism which leaves F( a, ) fixed. p (x) being separable, the roots a , a 2 , • • • , as of p (x) are a 1 1 distinct elements of E. By Theorem 8 there exist isomorphisms s. 45 au a 2 , •. , , a, mapping F ( a 1 ) on F ( a 1 ), F ( a 2 ), ••• , F (a,), respectively, which are each the identity on F and map u 1 on a 1 , a , ••• , as respectively. We now apply Theorem 10. E is a splitting 2 field of p(x) in F(a, ) and is also a splitting field of p(x) in F(a, ), Hence, the isomorphism ai, which makes p( x ) in F ( a ) correspond to 1 the same p(x) in F( a., ), can be extended to an isomorphic mapping of E onto E, that is, to an automorphism of E that we denote again by ai. Hence, a 1 , a 2 ,. • , as are automorphisms of E that leave F fixed and map a 1 ontO al'a , •.. a,. 2 Now let 8 be an element that remains fixed under all automor· phisms of E that leave F fixed. We know already that it is in F (a 1 ) and hence has the form e = co + c1a 1 -t c, a,2 -l . . . + c s-1 a 1s-1 where the ci are in F. If we apply a 1 to this equation we get, since a,(O)=O: The polynomial c s- 1 xs-l + c s-2 x has therefore the s distinct roots a 1 , a 2 s-2 ,. + + c x 1 , + (c 0 as. These are more than its degree. SJ all coefficients of it must vanish, among them c This shows 8 in e) 0 - e. F. Necessity. If E is a normal extension of F, then E is spiitting field of a separable polynomial p(x). We first prove the Lemma. If E is a normal extension of F, then E is a separable extension of F. Moreover any element of Eisa root of an equation over F which splits completely in E. 46 Let a 1 , a 2 , . . . , a~ be the group G of automo:rphisms of E whose fixed field is F. Let a be an elementofE, and let a, a 2 ,a 3 , the set of distinct elements in the sequence a 1 ( ... ,ar be a), a 2 ( a), ... , a 5 ( a). Since G is a group, a;(a,) = a;<a.(a)) = a;ak(a) = a,(u) = a,. Therefore, the elements a,a 2 , ... , ar are permuted by the automo:rphisms of G. The coefficients of the polynomial f(x) = ( x-a) ( x-u 2) • . . ( x-a,) are left fixed by each automorphism of G, since in its factored form the factors of :flx) are only permuted. Since the only elements of E which are left fixed by all the automorphisms of G belong to F, f(x) is a polynomial in F. If g(x) is a polynomial in F which also has a as root, then applying the automorphisms of G to the expression g (a ) = 0 we obtain g( a) = 0, oo that the degree of g(x) > s. Hence f(x) is irre- ducible, and the lemma is established. To complete the proof of the theorem, let cu 1 , w 2 , ... , w t be a gen- erating system for the vector space E over F. Let f 1( x) be the separable polynomial having w i as a root. Then E is the splitting field of p(x) = f 1 (x)-f 2 (x)· .... f,(x). If f(x) is a polynomial in a field F, and E the splitting field of f (x ), then we shall call the group of automo:rphisms of E over F the group of the equation f(x) = 0. We come now to a theorem known in algebra as the Fundamental Theorem of Galois Theory which gives the relation between the structure of a splitting field and its group of automo:rphisms. THEXJREM 16. (Fundamental Theorem). If p(x) is a separable polynomial in a field F, and G the group of the equation p(x) = 0 where E is the 47 splitting field of p(x), then: (1) Each intermediate field, B,is the fixed field for a subgroup G 8 of G, and distinct subgroups have distinct fixed fields. We say B and G 8 "belong" to each other. (2) The intermediate field B is a normal extension of F if and only if the sub- group G8 is a normal subgroup of G. In this case the group of automorphisms of B which leaves F fixed is isomorphic to the factor group ( G/G ). (3) For each intermediate field B, we have ( B/F) = index of G 8 and ( E/B) = order of G,. The first pm1 of the theorem cornes from the observation that E is splitting field for p(x) when p(x) is taken to be in any intermediate field. Hence, E is a normal extension of each intermediate field B, so that B is the fixed field of the subgroup of G consisting of the automorphisms which leave B fixed. That distinct subgroups have distinct fixed fields is stated in Corollary 2 to Theorem 14. Let B be any intermediate field. Since B is the fixed field for the subgroup G8 of G, by Theorem 14 we have ( E/B) = order of G,. Let us call o(G) the order of a group G and i(G) its index. Then o(G) "o(G 8 ) ·i(G 8 ). But (E/F) = o(G), and (E/F) = (E/B)·(B/F) from which ( B/F) " i (G, ), which proves the third part of the theorem. The number i( G 8 ) is equal to the number of lefr cosets of G,. The elements of G, being automorphisms of E, are isomorphisms of B; that is, they map B isomorphically into some other subfield of E and are the identity on F. The elements of G in any one coset of GB map B in the same way. For let aG 8 • Since a and a 1 2 a· a1 and a· a 2 be two elements of the coset leave B fixed, for each a in B 48 we have aa 1 ( a)= a( a)= aa 2 ( a). Elements of different cosets give different isomorphisms, for if a and r give the same isomorphism, o(a) = r(a) for each a in B, then a· 1 r(a):::: a for each a in B. Hence, 1 0 - , ::: a 1 , where a is an element of G rG 9 " aa 1 G 9 ~ aG 9 1 8 . But then r:::: aa 1 and so that a and r belong to the same coset. Each isomorphism of 8 which is the identity on F is given by an automorphism belonging to G. For let a be an isomorphism mapping 8 on B' and the identity on F. Then under a, p ( x) corresponds to p ( x ), and E is the splitting field of p ( x) in 8 and of p( x) in 8 ' , By Theorem 10, a can be extended to an automorphism a' of E, and since a I leaves F fixed it belongs to G. Therefore, the number of distinct isomorphisms of 8 is equal to the number of cosets of G 8 and is there- fore equal to ( B/F). The field a8 onto which a maps 8 has obviously aG 9 a-l as corresponding group, since the elements of a8 are left invariant by precisely this group. If B is a normal extension of F, the number of distinct automor- phisms of 8 which leave F fixed is ( B/F) by Theorem 14. Conversely, if the number of automorphisms is ( B/F) then B is a normal extension, because if F 1 is the fixed field of all these automorphisms, then F C F' C B, and by Theorem 14, ( B/F ')is equal to the number of automorphisms in the group, hence ( 8/F ') = ( 8/F ). From ( B/F) = (B/F')(F'/F) we have (F'/F) = 1 or F = F'. Thus, B is a normal extension of F if and only if the number of automorphisms of B is ( B/F ). B is a normal extension of F if and only if each isomorphism of 8 into E is an automorphism of B. This follows from the fact that each of the above conditions are equivalent to the assertion that there are 49 the same numberof isomorphisms and automorphisms. Since, for each a, B = aB is equivalent to aG 8 a -l C G 8 , we can finally say that B is a normal extension of F and only if G 8 is a normal subgroup of G. As we have shown, each isomorphism of B is described by the effect of the elements of some left coset of G,. If B is a normal exten- sion these isomorphisms are all automorphisms, but in this case the cosets are elements of the factor group ( G/G 8 ). Thus, each automor- phism of B corresponds uniquely to an element of ( G/G 8 versely. Since multiplication in ( G/G 8 ) ) and con- is obtained by iterating the mappings, the correspondence is an isomorphism between ( G/GB ) and the group of automorphisms of B which leave F fixed. This completes the proof of Theorem 16. I. Finite Fields. It is frequently necessary to know the nature of a finite subset of a field which under multiplication in the field is a group. The answer to this question is particularly simple. THEOREM 17. If S is a finite subset (f 0 ) of a field F which is a group under multiplication in F, then S is a cyclic group. The proof is based on the following lemmas for abelian groups. f.emma 1. If in an abelian group A and B are two elements of orders a and b, and if c is the least common multiple of a and b, then there is an element C of order c in the group. 50 Proof: (a) If a and b are relatively prime, C = AB has the required order ab. The order of C a = Ba is b and therefore c is divisible by b. Similarly it is divisible by a. Since Cab = 1 it follows c = ab. (b) If d is a divisor of a, we can find in the group an element of order d. Indeed Aa/d is this element. (c) Now let us consider the general case. Let pl' p 2 , ... , Pr be the prime numbers dividing either a or b and let n · · · Pr r Call ti the larger of the two numbers ni and c = P1 t 1 P2 t2 11\. Then tr, ··· P, · n- According to (b) we can find in the group an element of order pi 1 and . one of order pi one of order pi~ . Thus there 1s '• . Part (a) shows that the product of these elements will have the desired order c. _Lemma 2:_. If there is an element C in an abelian group whose _?rder c i~ maximal (as is always the case if the group is finite) then c is divisible by the order a of every element A in the group; hence ~c 1 is_ satisfied by each element in the group. Proof: If a does not divide c, the greatest common multiple of a and c would be larger than c and we could find an element of that order, thus contradicting the choice of c. We now prove Theorem 17. Let n be the order of S and r the largest order occuring in S. Then xr - 1 ::;:: 0 is satisfied for all ele- 51 ments of S. Since this polynomial of degree r in the field cannot have more than r roots, it follows that r :2 n. On the other hand r _s n be- cause the order of each element divides n. S is therefore a cyclic group consisting of 1, f, ( 2,, . where fn = 1. , fn-1 Theorem 17 could also have been based on the decomposition finite theorem for abelian groups having a number of generators. Since this theorem will be needed later, we interpolate a proof of it here. Let G be an abelian group, with group operation written as t-. The element g , . . , 1 gk will be said to generate G if each element g of , gk, g = n 1 g1 + G can be written as sum of multiples of g, If no set of fewer than k elements generate G, then g, called a minimal generating system. Any . + nkgk. , ~ will be group having a finite genera- ting system admits a minimal generating system. In particular, a finite group aLways admits a minimal generating system. From the identity n 1 ( g 1 it follows that if g 1 , g 2 , g,2, , .. gk 1 + mg,) + (n 2 .. n 1 m)g 2 = n 1g 1 + n 2 g 2 , gk generate G, also g 1 + mg, generate G. An equation m1 g 1 + m 2 g 2 + + mkgk = 0 will be called a re- lation between the generators, and m 1 , . . , mk will be called the co- efficients in the relation. We shall say that the abelian group G is the direct product of its subgroups G 1 , G ,. 2 sumg = x 1 + x2 t . , Gk if each g f G is uniquely representable as a + x k' where xi f G , i = 1, , , . , k. 1 52 Decomposition Theorem. Each abelian group having a finite number of generators is the direct product of cyclic subgroups G 1 , where the order of G i divides the order of G i"'"l, i = 1, the number of elements in a minimal generating system. may each be infinite, in which case, to be O(GJIO(G. ' Gn . n-1 and n is Gr, Gr+1 , Gn precise, )fori= 1,2, ... ,r-2). + We assume the theorem true for all groups having minimal generating systems of k-1 elements. If n = 1 the group is cyclic and the theorem trivial. Now suppose G is an abelian group having a minimal generating system of k elements. If no minimal generating system fies a non-trivial relation, then let g , g , 1 2 satis- , gk be a minimal generating system and G , G , . . . , Gk be the cyclic groups generated by them. 1 For each g t 2 G, g = n g + 1 1 + nkgk where the expression is unique; otherwise we should obtain a relation. Thus the theorem would be true. Assume now that some non-trivial relations hold for some minimal generating systems. Among all relations between minimal generating systems, let (1) be a relation in which the smallest positive coefficient occurs. After an eventual reordering of the generators we can suppose this coefficient to be m 1 , In any other relation between g , 1 , g,. (2) we must have m /n 1 1 • Otherwise n 1 = qm 1 + r, 0 < r < m1 and q times relation (1) subtracted from relation (2) would yield a relation with a coefficient r < m1 . Also in relation (1) we must have m/mi' i = 2,. , k. 53 For suppose m 1 m qm 2 1 + 1 <r < r, 0 g1 + g2 , g2 , ID 1( g does not divide one coefficient, say m2 ••• , rn . In the generating system 1 gk we should have a relation + qg 2 ) + rg 2 + m3 g 3 + r contradicts the Then choice of m 1 • + mkqk = 0 where the coefficient Hence q2 m1 , m2 = m3 = q3m 1 , , gk is minimal generating, and m1 g-1 = 0. In any relation 0 = n 1if1 + n 2 g 2 + g1 , since m 1 is a coefficient in a relation between vious argument yields m1 n 1 , and hence n 1 g·1 = g 2, g1 of G1 and G'. Each (n 1 - n;Jg1 -r- (g'- )g1 By our inductive hypothesis, G =0, so that n1 [ 1 n;g 1 = cyclic groups generated by elements . . . , tk . , gk . and G1 the Then G is the direct product satisfy ti titl, g1 + g1 = n •g1 1 + g" implies g") = 0, hence n; orders t 2 , gk our pre- element g of G can be written The representation is unique, since n 1 the relation (n 1 nkgk 0. Let G 1 be the subgroup of G generated by g2 , cyclic group of order m generated by 1 + l andalso g'= is the direct product of k-1 g2, g3 , , i = 2, ment applied to the generators g 1 , g 2 , . g<~. gk whose respective k-1. The preceding argugk yields m 1 t 2 , from which the theorem follows. By a finite field is meant one having only a finite number of elements. Corollary. The non-zero elements of a finite field form a cyclic If a is an element of a field F, let us denote the n-fold of a, i.e,, 54 the element ofF obtained by adding a to itself n times, by na. It is ob- vious that n·(m·a)= (nm)·a and(n-a)(m·b) = nm·ab. If for one element a f. 0, there is an integer n such that n. a = 0 then n. b = 0 for each bin F, since n. b = (n.a)(a- 1 b) --= Q.a- 1 b = O.. Ifthere is a positive integer p such that p · a = 0 for each a in F, and if p is the smallest integer with this property, then F is said to have the charac- 1eristic_p. If no such positive integer exists then we say F has characteristic 0. The characteristic of a field is always a prime number. for if P= r.sthenpa = rs.a=r.(s.a).However, s.a=b ,fOil a f,O -f and r b -,_; 0 since both r and S are less than p, to the definition of the characteristic. If na 0 for a ~ 0, then p divides n, for n = qp + r where 0 <r < p so that pa and na = (qp + 0 contrary r)a = qpa + ra. Hence na = 0 implies ra = 0 and from the definition of the characteristic since r < p, we must have r = 0. If F is a finite field having q elements and E an extension of F , w , •. ,w n is 1 2 a basis of E over F, each element of E can be uniquely represented as such that (E/F) = n, then E has qn elements. For if a linear combination x 1 w 1 + x F. Since each xi ble choices of x 1 , 2 w2 + f- {iJ xnwn where the x 1 belong to can assume q values in F, there are qn distinct possi. . , X 0 and hence qn distinct elements of E. E is finite, hence, there is an element a of E so that E = F(a). (The non- zero elements of E form a cyclic group generated by a). If we denote by P _ [ 0, 1, 2 , ... 1 p-1] the set of multiples of the unit element in a field F of characteristic p, then P is a subfield of F having p distinct elements. In fact, P is isomorphic to the field of integers reduced mod p. If F is a finite field, then the degree of F over 55 Pis finite, say (F/P} = n, and F contains p 0 elements. In other words, the order of any finite field is a power of its characteristic. IfF and F' are two finite fields having the same order q, then by the preceding, they have the same characteristic since q is a power of the characteristic. The multiples of the unit in F and F 1 form two fields P and P 1 which are isomorphic. The non-zero elements ofF and F' form a group of order q-1 and, therefore, satisfy the equation xq-l 1 ~ 0. The fields F and F 1 are splitting fields of the equation x q-1 = 1 considered as lying in P and p 1 respectively. By Theorem 10, the isomorphism between P and P can be extended to an isomorphism between F and F ' . We have thus proved THEOREM 18. Two finite fields having the same number of elements are isomorphic. Differentiation. If f(x) = a o + a x + . . . + a nxn is a poly1 nomial in a field F, then we define f 1 ~a, + 2a 2 x + ... + nanx n-l The reader may readily verify that for each pair of polynomials f and g we have (f+g)'=f+g' ( f g)' fg' + gf (fn)' THEOREM 19. The polynomial f has repeated roots if and only if in the splitting field E the polynomials f and f' have a common root. This condition is eauivalent to the assertion that f and ft have a 56 common factor of degree greater than 0 in F. If a is a root of multiplicity k of f(x) then f ~ (x-a )"Q (x) where ~ Q(a) f 1= ( x-<t 0. This gives )kQ 1 ( , ) + k ( x-a) k- 1 Q ( , ) ~ ( x-a) k-1 [ ( ,_, ) Q' ( , ) + kQ ( x)]. If k > 1, then a is a root of f' of multiplicity at least k-1. If k ~ 1, then f'(x) ~ Q(x) + (x-a)Q'(x) and f(a) ~ Q(a) ~ 0. Thus, f and f' have a root a in common if and only if a is a root off of multiplicity greater than 1. Iff and f' have a root a in common then the irreducible polynomial in F having a as root divides both f and f' . Conversely, any root of a factor common to both f and f' is a root off and f ' . Corollary. If F is a field of characteristic 0 then each irreducible polynomial in F is separable. Suppose to the contrary that the irreducible polynomial f(x) has a root a of multiplicity greater than 1. Then, f 1 (x) is a polynomial which is not identically zero (its leading coefficient is a multiple of the leading coefficient of :flx) and is not zero since the characteristic is 0) and of degree 1 less than the degree of f(x). But a is also a root of f (x) which contradicts the irreducibility of f(x). J I_!oots_ of Unity. If F is a field having any characteristic p, and E the splitting field of the polynomial x "- 1 where p does not divide t:t, then we shall refer to E as the field generated out ofF by the adjunction of a primitive nth root of unity. The polynomial x n its derivative, nx n-l , 1 does not have repeated roots in E, since has only the root 0 and has, therefore, no roots 57 in common with x n- 1. Thus, E is a normal extension cf F. If tl' t: 2 ,,.,, € 0 are the roots of x n- 1 in E, they form a group under multiplication and by Theorem 17 this group will be cyclic. If 1 ,f, f 2, ... , f n-1 are the elements of the group, we shall call t a primitive n th root of unity. The smallest power of f which is 1 is then th, THEOREM 20. If E is the field generated from F by a primitive nth root of unity, then the group G of E over F is abelian for any n and cyclic if n is a prime number. We have E ~ F(E), Thus, if a and since the roots of X0 - is an integer 1 ::no "r f:. rare distinct elements of G, a(£) rootofx" -1 and, hence, a power off, Thus, f 1 are powers of f, < n. r(f). But rJ(r) is a a(t)= Moreover, ra(i)::: r( ~ 00 ):::: f "awhere "a (r(f)) "a::: "a= ar(f), Thus, nar == q,n 7 mod n. Thus, the mapping of a on no is a homomorphism of G into a multiplicative subgroup of the integers mod n. Since r ;1:. a implies r(f) ~ a(£), it follows that r f:- a implies "a ~ n, mod n. Hence, the homomorphism is an isomorphism. If n is a prime number, the multiplicative group of numbers forms a cyclic group. K. Noether Equations. If E is a field, and G = ( any set of elements x0 , x, , a, r, . in E will be said to provide a solution to Noether's equations if xa. a ( x 1 element x 0 . ) a group of automorphisms of E, ) ::: x = 0 then x = 0 for each r 1 for each a and r in G. If one 07 f G. As all values in G, and in the above equation x 07 T ::: traces G, ar assumes 0 when x in any solution of the Noether equations no element 0 xa = 0. Thus, = 0 unless the solution is completely trivial. We shall assume in the sequel that the 58 trivial solution has been excluded. THEOREM 21. The system x , x , ... is a solution to Noether's equations if and only if there exists an element Xa a in E, such that ala( a ) for each a. For any a, it is clear that xa =a/a( a) is a solution to the equations, since a!<>(a)·<>(a/r(a)) = a/a(a)·a(a)/OT(a) = a/m(a). Conversely, let x x , ... be a non-trivial solution. Since the "' r automorphisms a, r, ... are distinct they are linearly independent, and the equation xa 'U(Z) + x, r ( z) + . . . = 0 does not hold identically. Hence, there is an element a in E such that xa""(a) + x,r(a) + ... = a~ 0. Applying a(a) = Multiplying by a to a gives L a(x,).ar(a). uG 'lr gives Xa·a(a) = Replacing xa . a ( x, ) by k xaa(x,).m(a). uG 'trr and noting that ar assumes all values in G when r does, we have Xa •<J(a) = L x,r(a) = a nG so that xa = a!a(a). A solution to the Noether equations defines a mapping C of G into E, namely, C(a) = xa· If F is the fixed field of G, and the elements x lie in F, then C is a character of G. For 0 C(ar)= x01 = xa·a(x,) = xaxr= C(a)·C(r)sincea(x,) = x, if x 7 f F. Conversely, each character C of G in F provides a solution 59 to the Noether equations. Call C(o) xo. Then, since xr f F, we have a( xr) = Xr . Thus, Xa ·a ( x,) = xa · x, = C (a) · C ( r) = C( or) xar. We thereforehave, by combining this with Theorem 21, THEOREM 22. If G is over the group of the normal field E F, then for each character C of G into F there exists an element a in E such that C(a) = a/a(a) and, conversely, if a/a( a) is in F for each a, then C (a)= a/a( a ) is a character of G. If r is the least common multiple of the orders of elements of G, then a' E F. We have already shown all but the last sentence of Theorem 22. To prove this we need only show a(ar) =a' for a'/a(a') = each a f G. But (a/a(a))' = (C(a))' = C(a') = C(l) = 1. L. Kummer's Fields. IfF contains a primitive nth root of unity, any splitting field E of a polynomial ( xn - a i = 1,2,. . 1 )( xn - a ) . . . ( xn - a ) where a. ( F for 2 ' ' , r will be called a Kummer extension of F, or more briefly, a Kummer ..-field., If a field F contains a primitive nth root of unity, the number n is not divisible by the characteristic of F. Suppose, to the contrary, F has characteristic p and n :::: qp. Then yP _ 1 = (y - 1 )P since in the expansion of (y - 1 )P each coefficient other than the first and last is divisible by p and therefore is a multiple of the p-fnld of the unit of F and thus is equal to 0. Therefore xn - 1 = (x q)p - 1 = ( xq - 1 )P and X 0 - 1 cannot have more than q distinct roots. But we assumed that F has a primitive nth root of unity and 1, t, t 2, . . . , ( n-1 would be 60 n distinct roots of xn - 1. It follows that n is not divisible by the characteristic of F. For a Kummer field E, none of the factors x" - a 1, a f 0 has repeated roots since the derivative, nxn·l , has 1 only the root 0 and has therefore no roots in common with xn - ai. Therefore, the irreducible factors of xn - a, are separable, so that E is a normal extension of F. Let a 1 be a root of xn- ai in E. If tinct nth roots of unity in F, then roots of xn- ai' and E = F(a 1 , a 2 , ••• , f 1 , f 2 , . . . , t: aifl' ait , . . . , 2 0 are then dis- a 1t:n will ben distinct hence will be the roots of xn -a, so that ar). Let a and r be two automorphisms in the group G of E over F. For each ai' both a and r map ai on some other root of x" - ai. Thus r(ai) = f 17 a 1 and a(a) = f iO ai where fia and fir are nth roots of unity in the basic field F. It follows that r(a(a,)) = r(<,aa,) = '•ur(a) = 'ia'lT a,= a(r(a,). Since oand r are commutative over the generators of E, they commute over each element of E. Hence, G is commutative. If a c.1 aa., a 2 (a.) 1 1 fia 01 =f. 2 a., 10, f G, then a (a i) = 01 etc. Thus, u (a.) =a.1 for n.1 such that 1 = 1. Since the order of an n th root of unity is a divisor of n, we have ni a divisor of n and the least common multiple m of n 1 , n 2 , •.. , n r is adivisorofn. Since a"(u 1 ) = ai fori= 1,2, ... , r it follows thatm is the order of a. Hence, the order of each element of G is a divisor of n and, therefore, the least common multiple r of the orders of the elements of G is a divisor of n. If fn/r f is a primitive n1h root of unity, then is a primitive r th root of unity. These remarks can be summarized in the following. 61 _THEXJREM 23. If Eisa Kummer field, i.e., a splitting field of p(x) = (x"- a 1 )(x"- a 2 ) contains a primitive nth •.• (x"- a,) whete a, lie in F, and F root of unity, then: (a) E is a normal extension ofF; (b) the group G of E over F is abelian, (c) the least common multi- ple of the orders of the elements of G is a divisor of n. Corollary._ If E is the splitting field of xP - a, and F contains a primitive p th root of unity where p is a prime number, then either E ::: F and xP -- a is split in F, or xP ... a is irreducible and the group of E over F is cyclic of order p. The order of each element of G is, by Theorem 23, a divisor of p and, hence, if the element is not the unit its order must be p. If root of xP - a, then a, m, . . . , £p-I a are all a is a the roots of xP .... a so that F(a ) = E and ( E/F) <]' . Hence, the order of G does not exceed p oo that if G has one element different from the unit, it and its powers must constitute all of G. Since G has p distinct elements and their behavior is determined by their effect on a, then a must have p distinct images. Hence, the irreducible equation in F for a must be of degree p and is therefore xP ... a ::: 0. The properties (a), (b) and (c) in Theorem 23 actually characterize Kummer fields. Let us suppose that E is a normal extension of a field F, whose group G over F is abelian. Let us further assume that F contains a primitive r th root of unity where r is the least common multiple of the orders of elements of G. The group of characters X of G into the group of r th roots of 62 unity is isomorphic to G. Moreover, to each a a character C f f G, if a f:- 1, there exists X such that C (a) /:. 1. Write G as the direct product of the cyclic groups G, G 2 , . m,. Each Gt of orders m 1 m 2 af G Call C i the character sending a i into ~',-, a primitive m . 'h root of unity and a. into 1 for j ~ i. Let C be • anyc h aracter. C ( ai ) Conversely, C J ::::£i 1 11 •,ten h we h ave C = C 1 '' C 1-lt fLi • t I defines . C I' vl Thus C a 1 ( 1/ 2 2 . , vt C t I'' i a character. Since the order of C. is • m 1, the character group X of G is isomorphic to G. If a= a- 1 2 a f:- 1, then in at least one vi, say v , is not divisible by m 1 . .a 1 1 a)~ , v jo 1. 1 1 Let A denote the set of those non-zero elements a of E for which a' c F and let F denote the non-zero elements of F. It is obvious that 1 A is a multiplicative group and that F1 is a subgroup of A. Let N denote the set of rth powers of elements in A and F~ the set of r th powers of elements of F 1 . The following theorem provides in most applications a convenient method for computing the group G. THEOREM 24. The factor groups (A/F ) and (A'/F ')are isomorphic to each other and to the groups G and X. We map A on A' by making a t A correspond to ar (A'. If ar f F~, where a E F then b E A is mapped on fl if and only if br = ar, that is, 1 if b is a solution to the equation xr ... ar = 0. But a, ca, c 2a, ... , (r-1 a are distinct solutions to this equation and since f and a belong to it follows that b must be one of these elements and must belong to Thus, the inverse set in A of the subgroup F~ of N is F 1 factor groups ( A/F 1 ) and ( N IF~ ) are isomorphic. , F1, F1 . so that the 63 If a is an element of A, then (a/a(a))r = ar/u(ar) = 1. Hence, a/a( a) is an root of unity and lies in F r'h 1 • ~Theorem 22, a/a (a) defines a character C (a) of Gin F. We map a on the corresponding character C. Each character C is by Theorem 22, image of some Moreover, a.a 1 is mapped on the character C a. * (a) = a· a'/a(a·a') = a·a'!a(a)·a(a') = C(a)·C'(a) = C.C'(a), so that the mapping is homomorphism. The kernel of this homomorphism is the set of those elements a for which a/u(a) = 1 for each is F 1 • It follows, therefore, that ( A/F 1 also to G. In particular, (A/F 1 ) ) 0 , hence is isomorphic to X and hence is a finite group. We now prove the equivalence between Kummer fields and fields satisfYing (a), (b) and (c) of Theorem 23. THEDREM 25. If E is an extension field over F, then E is a Kummet field if and only if E is normal, its group G is abelian and F contains a primitive rth root f of unity where r is the least common multiple of the orders of the elements of G. The necessity is already contained in Theorem 23.. We prove the sufficiency. Out of the group A, let a1 F 1 , a 2 F 1, ... , atF1 be the co sets ofF 1 . Since ai (A, we have af = ai (F. Thus, ai is a root of the equation xr- a.=Oand sincem.,c 2a., . . . , ' ' ' fr-la., are also roots, xr - ai must split in E. We prove that E is the splitting field of (x'- a 1 )(x'- a 2 ) •.. (x'- a,) which will complete the proof of the theorem. 'lb this end it suffices to show that F ( a , a , . . . , at) =E. 1 2 64 , a,) Suppose that F ( a , a , . 1 2 /c E. Then F(a, intermediate field between F and E, and since E is normal over F (a,, . , at) there exists an automorphism a c G, leaves F (a 1 , •.. , f- 1, which a,) fixed. There exists a character C of G for which a C(a) -/:. 1. Finally, there exists an element C(o) ~ a/a(a) ~ 1. But a' Moreover, A c F(a, in F(a, (1 f in E such that F 1 by Theorem 22, hence a, A. ,a, ) since all the cosets a iF1 are ,a,). Since F(a, contained , at) is by assumption left fixed by a, a( a) = a which contradicts a/a(a) f- 1. It follows, therefore, that F{a,, ... ,a 1 ) =E. Corollary. If E is a normal extension of F, of prime order p, and _ifF contains a primitive pth root of unity, then E is splitting field of an irreducible polynomial xP ... a in F. a1 , E is generated by elements , an where af not in F . Then xP - a is irreducible, for otherwise F(a 1 F. Let a 1 be would be an intermediate field between F and E of degree less than p, and by the product theorem for the degrees, p would not be a prime number, contrary to assumption. E :::; F ( a 1 ) is the splitting field of xP - a. M. Simple Extensions. We consider the question of determining under what conditions an extension field is generated by a single element, called a primitive. We prove the following THEOREM 26. A finite extension E of F is primitive over F if 65 ~x_j! there are only a finite number of intermediate fields. (a) Let E a in F. ~ F(a) and call f(x) ~ 0 the irreducible equation for Let B be an intermediate field and g(x) the irreducible equa- tion for a in B. The coefficients of g(x) adjoined to F will generate a field B 1 between F and B. g ( x ) is irreducible in B, hence also in B Since E = B'(a) we see ( E/B ) ~ ~ ( E/B ' ). This proves B' 1• B. So B is uniquely determined by the polynomial g(x). But g(x) is a divisor of :flx), and there are only a finite number of possible divisors of :flx) in E. Hence there are only a finite number of possible B's. (b) Assume there are only a finite number of fields between E and F. Should F consist only of a finite number of elements, then E is generated by one element according to the Corollary on page 53. We may therefore assume F to contain an infinity of elements. We prove: To any two elements a, (3 there is a y in E such that F (a, (3 ) ~ F (y). Let y :::a+ a(3 with a in F but for the moment undetermined. Con- sider all the fields F(y) obtained in this way. Since we have an infinity of a's at our disposal, we can find two, say a, and a 2 , such that the corresponding y's, y1 ~ the same field F ( y 1 ) ~ F ( y 2 ). a + aJ3 and y, ~ a + a2 (3, Since both y and y are in F ( y 1 2 yield 1 ), their difference (and therefore (3) is in this field. Consequently also y1 - a1 f3 ~a. So F(a,(3) C F(y 1 ). Since F(y 1 ) C F(a,(3) our con- tention is proved. Select now ~ in E in such a way that ( F ( as large as possible. Every element could find an element proves E ~ F(1). € ~ )IF ) is of E mustbe in F( 1J) or else we osuch that F(6) contains both ~and '·This 66 THEOREM 27. If E = F(a, a 2 , the field F, and a ~xists 1 a primitive , . . . , an) is a finite extension of a 2, ... , a 0 are separable elements in E, then there ein E such that E = F (e). Proof: Let fi ( x) be the irreducible equation of ai in F and let B be an extension of E that splits fi(x) 12 ( x) ... f,(x). Then B is normal over F and contains, therefore, only a finite number of intermediate fields (as many as there are subgroups of G). 8:) the subfield E contains only a finite number of intermediate fields. Theorem 26 now compietes the proof. N. Existence of a Normal Basis. The following theorem is true for any field though we prove it only in the case that F contains an infinity of elements. fHE_OREM 28. If E is a normal extension ofF and a 1 , a , ... , a are the elements of its group G, there is an element & in E such that then elements a,( e), a,( 0)' ... , an( 0) are linearly independent with respect to F. According to Theorem 27 there is an a such that E = F(u). Let f(x) be the equation for a, put o,(a) = a i, f(x) f(x) g(x) = (x-a)f'(a) and g,(x) = o,(g(x)) = -(x-q,)f'(q,) g i( x ) is a polynomial in E having ak as root fork (1) g /X) gk (X) = 0 (mod f(x)) for i I= i and hence t k. In the equation (2) g*(x) + g 2 (x) + ... + g,(x)- 1 = 0 the left side is of degree at most n - 1. If (2) is true for n different values of x, the left side must be identically 0. Such n values are 67 apa 2 , ••• ,a,, since gi( ai) = 1 and gk(ai) c:::O fork-/::. i. Multiplying (2) by g,( x) and using (I) shows: (g,(x))' (3) g,(x) (mod f(x)). = We next compute the determinant D(x) = 1a,ak(g(x)) I i,k = 1, 2, ... , n (4) and prove D(x) ~ 0. If we square it by multiplying column by column and compute its value (mod f(x)) we get from (1), (2), (3) a determinant that has 1 in the diagonal and 0 elsewhere. s0 (D(x)) 2 =I (mod f(x)). D (x ) can have only a finite number of roots in F. Avoiding them we can find a value a for x such that D(a) /:. 0. Now set (J = g(a). Then the determinant (5) Consider any linear relation t ing the automorphism xnan( ()) = 0 where the xi are in F. Apply- a. to it would lead to n homogeneous equations for ' the n unknowns xi. (5) shows that xi= 0 and our theorem is proved. 0. Theorem on Natural Irrationalities. Let F be a field, p(x) a polynomial in F whose irreducible factors are separable, and let E be a splitting field for p(x). Let B be an arbitrary extension of F, and let us denote by EB the splitting field of p(x) when p(x) is taken to lie in B. If rx 1 , EB, then F(a, . , a 6 are the roots of p(x) in ,as) is a subfield of EB which is readily seen to form a splitting field for p(x) in F. By Theorem 10, E and F(a 1 , . a 5 ) 68 are isomorphic. There is therefore no loss of generality if in the we take E ::: F ( a 1 , ••• , ofEB.Also,EB = B (a sequel a,) and assume therefore that E is a subfield 1 Let us denote by E , ... , n a,). B the intersection of E and B. It is readily seen that E n B is a field and is intermediate to F and E. THEOREM 29. If G is the group of automorphisms of E over F, _?.nd H th~ group of EB over B, then H is isomorphic to the subgroup of G having E n B as its fixed field. Each automorphism of EB over B simply permutes al' ... , a,. in some fashion and leaves B, and hence also F, fixed. Since the elements of EB are quotients of polynomial expressions in a , . . , as with 1 coefficients in B, the automorphism is completely determined by the permutation it effects on a 1 , • • , as, Thus, each automorphism of EB over B defines an automorphism of E = F ( a 1 , fixed. Distinct automorphisms, since a 1 , • .. , a 5 ) which leaves F , a 8 belong to E, have different effects on E. Thus, the group H of EB ovet B can be considered as a subgroup of the group G of E over F. Each element of H leaves E n B fixed since it leaves even all of B fixed. Howevet, any element of E which is not in E n B is not in B, and hence would be moved by at least one automorphism of H. It follows that E n B is the fixed field of H. ~orollaty:. If, undet the conditions of Theorem 29, the gtoup G is of _prime on!er, then either H = G or H consists of the unit element alone. 69 III APPLICATIONS by A. N. Milgram A. _Solvable_ Groups. Before proceeding with the applications we must discuss certain questions in the theory of groups. We shall assume several simple propositions: f(x) = xN (a) If N is a normal subgroup of the group G, then the mapping is a homomorphism of G on the factor group GIN. f is called the natural homomorphism. (b) The image and the inverse image of a normal subgroup under a homomorphism is a normal subgroup. (c) I f f is a homomorphism of the group G on G' , then setting N' = f(N), and defining the mapping g as g( xN ) = f(x) N a homomorphism of the factor group 1 , we readily see that g is G/N on the factor group G 1/N 1 • Indeed, if N is the inverse image of N' then g is an isomorphism. We now prove THEOREM 1. (Zassenhaus). If U and V are subgroups of G, U and v normal subgroups of U and V, respectively, then the following three factor groups are isomorphic: u (U nV) /u (U nv), v(UnV)/v(unV), (UnV)/(unV)(vnU). It is obvious that U n v is a normal subgroup of U n V. Let f be the natural mapping of U on U/u. Call f(UnV) = H and f(Unv) = K. Then f-'(H) = u(UnV) and C 1(K) = u(Unv) from which it follows that u( UnV)/u( Unv) is isomorphic to H/K. If, however, we view f as defined only over U n V, then f-\K) = [un(UnV)](Unv) = (unV)(Urw) so that (UnV)/(unV)(Unv) is also isomorphic to H/K. 70 Thus the first and third of the above factor groups are isomorphic to each other. Similarly, the second and third factor groups are isomorphic. Corollary 1. If H is a subgroup and N a normal subgroup of the group G, then H/HnN is isomorphic to HN/N, a subgroup of G/N. Proof: Set G = U, N = u, H = V and the identity 1 :::: v in Theorem 1. Corollary 2. Under the conditions of Corollary 1, if GIN is _abelian, _so also is H/HnN. Let us call a group G solvable if it contains a sequence of subgroups G = G0 _) G1 .J, .. ) Gs = 1, each a normal subgroup of the preceding, and with G i-1 ./G.1 abelian. THEOREM 2. Any subgroup of a solvable group is solvable. For let H be a subgroup of G, and call H 1 = HnG 1• Then that Hi_/Hi is abelian follows from Corollary 2 above, where Gi-l' Gi and Hi-l play the role of G, N and H. THEOREM 3. The homomorph of a solvable group is solvable. Let f(G) = G' , and defineG 1 i =f( Gi)where Gi belongs to a a sequence exhibiting the solvability of G. Then by (c) there exists a homomorphism mappingGi_/Gi on G~_/Gf. But the homomorphic image of an abelian group is abelian so that the groups G ~ exhibit the solvability of G' B. _Permutation Groups. Any one to one mapping of a set of n objects on itself is called a permu_tation: The iteration of two such mapping is called their product. 71 It may be readily verified that the set of all such mappings forms a group in which the unit is the identity map. The group is called the symmetric group on n letters. Let us for simplicity denote the set of n objects by the numbers L 2, ... , n. The mapping S such that S(i) noted by (123 ... n) and more generally (i j ping T such that T ( i) i + 1 mod n be de- . m) will denote the map- j, ... , T ( m) ~ i. If ( i j ... m) has k numbers, then it will be called a k cycle. It is clear that if T = (i j T- 1 will s) then = (s . . ji). We now establish the Lemm.::: If a subgroup U of the symmetric group on n letters (n > 4) contains every 3-cycle, and if u is a normal subgroup of U such that U/u is abelian, then U contains every 3-cycle. Proof: Let f be the natural homomorphism f(U) ,.... U/u and let X = (ijk), y (krs) be twoelements of U, where i, j, k, r, s are 5 numbers. Then since U/u is abelian, setting f(x) = X • , f(y) = y we have 1 f ( x-ly-1 xy) = X •-1 y H X 1 y 1 = 1, so that x- 1y- 1 xy ( u. But x-ly-1xy= (kji)·(srk)·(ijk)·(krs) = (kjs) and for each k, j, s we have (kjs) : u. THEOREM 4. The symmetric group G on n letters is not solvable for n > 4. If there were a sequence exhibiting the solvability, since G con- tains every 3-cycle, so would each succeeding group, and the sequence could not end with the unit. 72 C. Solutjon of Equations by Radicals. The extension field E over F is called an extension by radicals if there exist intermediate fields 8 1 , B2 , ... , Br = E and Bi = Bi_ 1( a) where each ai is a root of an equation of the form x ni - a. "" 0, ' a 1 ' Bi-1 . A polynomial f ( x ) in a field F is said to be solvable by radicals if its splitting field lies in an extension by radicals. We assume unless otherwise specified that the base field has characteristic 0 and that F contains as many roots of unity as are needed to make our subsequent statements valid. Let us remark first that any extension ofF by radicals can always be extended to an extension ofF by radicals which is normal over F. Indeed B is a normal extension of B 0 since it contains not only a , 1 butm 1 , 1 where (is any n 1 -root of unity, so that 8 1 is the splitting field of x"t- a,. If f (x) ~ TT(x"2- a( a )), where a takes all values in 1 2 (J the group of automorphisms of B 1 over B 0 , then f 1 is in B 0 , and joining successively the roots of xn 2 - a( a 2 ) ad- brings us to an exten- sion of B which is normal over F. Continuing in this way we arrive at 2 an extension of E by radicals which will be normal over F. We now prove THEOREM 5. The polynomial f(x) is solvable by radicals if and only if its group is solvable. Suppose f(x) is solvable by radicals. Let E be a normal extension of F by radicals containing the sphtting field B of f(x), and call G the group of E over F. Since for each i, Bi is a Kummer extension of Bi-1, the group of Bi over Bi-t is abelian. In the sequence of groups 73 = Gil =' = J ... :J G 8 , 1 each is a normal subgroup of the 1 preceding since GB is the group of E over Bi·l and Bi is a normal G G8 i-1 extension of Bi-l. But GB _{G 8 . /s the group of (\ over B1_1 and hence 1 is abelian. Thus G is solvable. However, GB is a normal subgroup of G, and G/GE is the group of B over F, and is therefore the group of the polynomial f(x). But G/G 8 is a homomorph of the solvable group G and hence is itself solvable. On the other hand, suppose the group G of f(x) to be solvable and let E be the splitting field. Let G = G 0 J G 1 :J . . . :J G, = 1 be a sequence with abelian factor groups. Call B 1 the fixed field for G 1 • Since Gi-l is the group of E over B 1_1 and Gi is a normal subgroup of Gi·l' then Bi is nmmal over B1_1 and the group Gi_/Gi is abelian. Thus Bi is a Kummer extension of Bi-t' hence is splitting field of a polynomial of the fonn ( x"~• 1 )( x"-a 2 ) ••• ( splitting fields of the x n - ak x"-a,) so that by forming the successive we see that Bi is an extension of B 1_1 by radicals, from which it follows that E is an extension by radicals. Remark. The assumption that F contains roots of unity is not necessary in the above theorem. For if f(x) has a solvable group G, then we may adjoin to F a primitive nth root of unity, where n is, say, equal to the order of G. The group of f(x) when considered as lying in F' is, by the theorem on Natural hrationalities, a subgroup G 1 of G, and hence is solvable. Thus the splitting field over F 1 of f(x) can be obtained by radicals. Conversely, if the splitting field E over F of f(x) can be obtained by radicals, then by adjoining a suitable root of unity E is extended to E 1 which is still normal over F 1 • But E 1 could be 74 obtained by adjoining first the root of unity, and then the radicals, to F; F would first be extended to F E 1 • 1 Calling G the group of E we see that G 1 1 and then F ' would be extended to over F and G is solvable and G/G 1 1 1 the group of E is the group ofF 1 over F 1 , over F and hence abelian. Thus G is solvable. The factor group G/GE is the group of f(x) and being a homomorph of a solvable group is also solvable. D. The General Equation of Degree n. If F is a field, the collection of rational expressions in the variables ul' u 2 , ... , U0 with coefficients in F is a field F ( u 1 , u2 , . . . , 11 0 By the general equation of degree n we mean the equation f(x) = x"- u 1xn-l + u 2x"- 2 - + . . . + ( -1 )"u•. (1) Let E be the splitting field of f(x) over F(u 1 , u 2 , v 1 , v .2, ul ••. , V 0 un). If are the roots of f(x) in E, then = VI + v2 + . . . + vn, u2 = vl v2 + vl v3. + tVn-lVn''' We shall prove that the group of E over F ( u 1 , u 2 , symmetric •.. , ... , U 0 ) is the group. Let F ( x,, x 2 , ••• , xn) be the field generated from F by the variables xl, x2, ...• xn. Let al = xt + x2 + ... + xn, a2 = x 1 x2 + x 1 x 3 -· .. , + X 0 _ 1 x0 , ... ,an:: x 1 x 2 • . • xn be the ele- mentary symmetric functions, i.e., (x-x ,)(x-x 2 ). . xn- a xn-I +- ... (-l)"an = f*(x). If g(a 1 ,a 2 , . 1 polynomial in a 1 , ... ,a,, then g(a 1 ,a 2 , ... (x-x,) .,a,) is a ,a,)= 0 only if g is the ). 75 Ixi, Lx 1xk, zero polynomial. For if g( , ) = 0, then this relation would hold also if the xi were replaced by the vi. Thus, g(Lvi, Lvi 0 or g(u 1 , u 2 , = vk, ... ) ... , un) = Ofromwhichitfollows that g is identically zero. Between the subfield F(a, F(u 1 , u 2 , •.. , U 0 we set up the following correspondence: Let ) f(u 1 , , , , , U 0 )/g(u 1 ,, f (a 1 , this correspond to ping of F ( u f(a 1 ,a 2 , •• , u , 1 2 1 U0 , • . )beanelementofF(u 1 , , a0 ) / g (a 1' Un) on all of F ( a 1 , , ,a 0 )/g(a 1 ,a 2 , ••• •. ,U 0 ).Wemake ,a,). This is clearly a map,a,). Moreover, if ,an) 0. But this , an), then fg, - gf, = £1 (at,a:!• ... ,an)/gl(al,a2'. implies by the above that f(u 1 , ... ,un)·g 1 (u 1 , ..• ,un)-g(u 1 , so that f(ut' ... , un)/g(ut' u 2 , = f ( U , .•• , 1 1 li 0 ) / g 1( •.. , , ... ,un).f 1 (u 1 , .. ,un)= 0 u.) U , U , .•. , lln ). 1 2 themappingof F(u 1 ,u 2 ... It follows readily from this that ,un) on F(a 1 ,f1 2 , . . . ,an)is an isomor- phism. But under this correspondence f(x) corresponds to f * ( x). Since E and F(x 1 , x 2 , .•• , xn) are respectively splitting fields of f(x) and f * ( x ), by Theorem 10 the isomorphism can be extended to an iso- morphism between E and F ( x 1 , x , . . , xn ). Therefore, the group of E 2 over F( u 1 , u , ... , un) is isomorphic to the group ofF ( x , x , ... , xn) 2 1 2 over F(a 1 ,a 2 , •. ,a,). Each permutation of xl' x 2 ,, .. , xn leaves al'a 2 , and, therefore, , an fixed induces an automorphism ofF( XI' x 2 , . leaves F(a 1,a , ... , a,.,) fixed. Conversely, each automorphism of 2 F ( x , x , . , , , xn ) which leaves F(a 1 , . . , a, ) fixed must permute the 1 2 roots x , x,., ... , x of f*(x) and is completely determined by the 1 " n 76 permutation it effects on x 1 , x 2 , , , • 1 xn. Thus, the group of F ( x 1 , x 2 , ... , xn) over F(a 1 ,a 2 , ... ,an) is the symmetric group on n letters. Because of the isomorphism between F ( x 1 , F ( u 1 , u2 , .•• , , , , 1 xn ) and E, the group for E over un) is also the symmetric group. If we remark that the symmetric group for n > 4 is not solvable, we obtain from the theorem on solvability of equations the famous theorem of Abel: THEOREM 6. The group of the general equation of degree n is the symmetric group on n letters. The general eguation of degree n is not solvable by radicals if n > 4. E. Solvable Equations of Prime Degree. The group of an equation can always be considered as a permu- tation group. If f (X) is a polynomial in a field F, let al' a,' . ' '' an be the roots of f(x) in the splitting field E ~ F( a 1 , , . , , a 0 ), Then each automorphism of E over F maps each root of f(x) into a root of f(x), that is, permutes the roots. Since E is generated by the roots of f ( x ), different automorphisms must effect distinct permutations. Thus, the group of E over F is a permutation group acting on the roots a 1 ,a 2 , ... ,a, of f(x). For an irreducible equation this group is always transitive. For let a and a' be any two roots of f(x), where f(x) is assumed irreduci- ble. F(a) and F(a ')are isomorphic where the isomorphism is the identity on F, and this isomorphism can be extended to an automorphism of E (Theorem 10). Thus, there is an automorphism sending any given root into any other root, which establishes the "transitivityn of the group. 77 A permutation ~tion_modulo that o(i) = bi a of the numbers 1, 2, , .. , q is called a linear q if there exists a number b ~ 0 modulo q such + c(mod q), i = 1, 2, . . . ,q. JHEOREM 7. Let f( x ) be an irreducible equation of prime de~ in a field F. The group G of f( x) (which is a permutation group of the roots, or the numbers 1, 2, ... , q) is solvable if and only if, after a suitable change in the numbering of the roots, G is a group of linear substitutions modulo q, and in the group G all the substitutions withb = 1, a(i) = c + l(c = 1, 2 , ... , q) occur. Let G be a transitive substitution group on the numbers 1, 2, ... , q and let G 1 be a normal subgroup of G. Let 1, 2, ... , k be the images of 1 under the permutations of G 1 ; we say: 1, 2, ... , k is a domain of transitivity of G,. If i <_q is a number not belonging to this domain of transitivity, there is a a f G which maps 1 on i. Then a{ 1, 2, ... , k) is a domain of transitivity of aG 1 a -1 . Since G1 is a normal subgroup of G, we have G 1 ::: aG 1a· 1 • Thus, u( 1, 2, ... , k) is again a domain of transitivity of G 1 which contains the integer i and has k elements. Since i was arbitrary, the domains of transitivity of G 1 all contain k elements. Thus, the numbers 1, 2, ... , q are divided into a collection of mutually exclusive sets, each containing k elements, so that k is a divisor of q. Thus, in case q is a prime, either k = 1 (and then G 1 consists of the unit alone) or k :::: q and G 1 is also transitive. To prove the theorem, we consider the case in which G is solvable. Let G = G 0 "J G 1 "J . . . J G s+l ~ 1 be a sequence exhibiting the solvability. Since Gs is abelian, choosing a cyclic subgroup of it 78 would permit us to assume the term before the last to be cyclic, i.e., Gs is cyclic. If a is a generator of Gs, a must consist of a cycle con- taining all q of the numbers 1, 2,,. , , q since in any other case G, would not be transitive [if o = ( lij ... m)( n .. , p).,. then the powers of a would map 1 only into 1, i, j , , .. m, contradicting the transitivity of G, ]. By a change in the number of the permutation letters, we may assume a ( i) ~ i + 1 (mod q) a"(i) =o i + c (mod q) Now let r be any element of G , TaT -1 is an element of G,, say of G,. 1 then rar- 1( j) = ab( j) =j + b (mod 5 _ 1 . Since Gs is a normal subgroup TaT - 1 ~ a h. Let r( i) = j or ,- 1 ( j) = q). Therefore, ra ( i) oc r ( i) ~ b ( mod q) or r (i tl ) setting r(O) ~ c, we have r(l) =r ( i) =c + + b for each i. Thus, b, r(2) =or( 1) + b = c + 2b and in general r(i):: c-'- ib (mod q). Thus, each substitution in G s~l is a linear substitution. Moreover, the only elements of leave no element fixed belong toG , since for each a i such that ai +b ' =; Gs-l ~ i (mod q) [take i such that (a-1) i which 1, there is an =- b]. We prove by an induction that the elements of G are all linear substitutions, and that the only cycles of q letters belong toG,. Suppose the assertion true of Gs-n. Let r f Gs-n-l and let a be a cycle which belongs to Gs (hence also to Gs-n ). Since the transform of a cycle is a cycle, r -lOT is a cycle in G!1-n and hence belongs to Gn , Thus ,- 1m ::. ab for some b. By the argument in the preceding para- graph, r is a linear substitution hi + c and if T itself does not belong to Gs, then r leaves one integer fixed and hence is not a cycle of q elements. 1, 79 We now prove the second half of the theorem. Suppose G is a group of linear substitutions which c(i) :::= contains a subgroup N of the form i + c. Since the only linear substitutions which do not leave an integer fixed belong to N, and since the transform of a cycle of q elements is again In each coset N where a ::_ i a cycle of q elements, N is a normal subgroup of G. r where r(i) ="- bi + c. But a- 1 r( r(i) o-bi and r• (i) i) =- b'i then = (bi + c the substitution a- 1 r occurs, + c) - c ::_ bi. Moreover, if rr 1 (i) ~ bb'i. Thus, thefactorgroup ( G/N) is isomorphic to a multiplicative subgroup of the numbers 1,2, ... , q-1 mod q and is therefore abelian. Since (G/N) and N are both abelian, G is solvable. Corollary 1. If G is a solvable transitive substitution group on q • • letters (q prime), then the only substitution of G which leaves two or more letters fixed is the identity. This follows from the q and bi + c =i fact that each substitution is linear modulo (mod q) has either no solution (b :;::; 1, c exactly one solution(b 1=- 1) unless b = 1, c =- f. 0) or 0 in which case the sub- stitution is the identity. Corollary 2. A solvable, irreducible equation of prime degree in a field which is a subset of the real numbers has either one real root or all its roots are real. The group of the equation is a solvable transitive substitution group on q (prime) letters. In the splitting field (contained in the field of complex numbers) the automorphism which maps a number into its complex conjugate would leave fixed all the real numbers. By Corollary 80 1, if two roots are left fixed, then all the roots are left fixed, so that if the equation has two real roots all its roots are real. F. Ruler and Compass Constructions. Suppose there is given in the plane a finite number of elementary geometric figures, that is, points, straight lines and circles. We seek to construct others which satisfy ce11ain conditions in terms of the given figures. Permissible steps in the construction will entail the choice of an arbitrary point interior to a given region, drawing a line through two points and a circle with given center and radius, and finally intersecting pairs of lines, or circles, or a line and circle. Since a straight line, or a line segment, or a circle is determined by two points, we can consider ruler and compass constructions as con- structions of points from given points, subject to certain conditions. If we are given two points we may join them by a line, erect a perpendicular to this line at, say 1 one of the points and, taking the distance between the two points to be the unit, we can with the compass lay off any integer n on each of the lines. Moreover, by the usual method, we can draw parallels and can construct m/n. Using the two lines as axes of a cartesian coordinate system, we can with ruler and compass construct all points with rational coordinates. If a, b, c, ... are numbers involved as coordinates of points which determine the figures given, then the sum, product, difference and quotient of any two of these numbers can be constructed. Thus, each 81 element of the field R( a, b, c, ... ) which they generate out of the rational numbers can be constructed. It is required that an arbitrary point is any point of a given region. If a construction by ruler and compass is possible, we can always choose our arbitrary points as points having rational coordinates. If we join two points with coefficients in R( a, b, c, ... ) by a line, its equation will have coefficients in R( a, b, c, . . . ) and the intersection of two such lines will be a point with coordinates in R( a, b, c, ... ). The equation of a circle will have coefficients in the field if the circle passes through three points whose coordinates are in the field or if its center and one point have coordinates in the field. However, the coordinates of the intersection of two such circles, or a straight line and circle, will involve square roots. It follows that if a point can be constructed with a ruler and com- pass, its coordinates must be obtainable from R( a, b, c, ... ) by a formula only involving square roots, that is, its coordinates will lie in a field R, J R,_ 1 field over ) ... Ri-l J R 1 = R(a, b, c, ... ) where each field R 1 is splitting of a quadratic equation x 2 - 6, p. 21) since either R, = R 1_1 or ( R/R 1_1 a :::: 0. It follows (Theorem ) = 2, that ( R,IR 1 ) is a power of two. If x is the coordinate of a constructed point, then (R 1 ( x)/R 1 ) •( R,/R 1 (x)) = (R,/R,) = zv oo that R 1 ( x)/R 1 must also be a power of two. Conversely, if the coordinates of a point can be obtained from R( a, b, c, ... ) by a formula involving square roots only, then the point can be constructed by ruler and compass. For, the field operations of 82 addition, subtraction, multiplication and division may be performed by ruler and compass constructions and, also, square roots using 1: r = r : r to obtain r ::: 1 compass v1 r 1 may be performed by means of ruler and instructions. As an illustration of these considerations 1 let us show that it is impossible to trisect an angle of 604 Suppose we have drawn the unit circle with center at the vertex of the angle, and set up our coordinate system with X-axis as a side of the angle and origin at the vertex. Trisection of the angle would be equivalent to the construction of the point (cos 20°, sin 20°) on the unit circle. From the equation cos 30 = 4 cos3 0 - 3 cos 0, the abscissa would satisfy 4x 3 - 3x = 1/2. The reader may readily verify that this equation has no rational roots, and is therefore irreducible in the field of rational numbers. But since we may assume only a straight line and unit length given, and since the 60° angle can be R (a, b, c,. . constructed, we may take ..) to be the field R of rational numbers. A root a of the irreducible equation 8x3 - 6x - 1 ~ o is such that ( R (a )/R) ~ 3, and not a power of two.